Coordinate Geometry · Class 10 Mathematics: 171 MCQs with answers

All 171 multiple-choice questions from this kit, free and with no sign-up.

Warm-up · 47 questions

Q1. The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is

Beginner
A $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ B $(x_2-x_1)^2 + (y_2-y_1)^2$ C $\sqrt{(x_2-x_1)^2 - (y_2-y_1)^2}$ D $\sqrt{x_2-x_1} + \sqrt{y_2-y_1}$
Show answer & explanation

$\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

The differences are squared, added, then rooted.

Q2. The distance formula comes from

Beginner
A the angle sum property B the section formula C the quadratic formula D Pythagoras' theorem
Show answer & explanation

Pythagoras' theorem

The two coordinate gaps are the legs of a right triangle.

Q3. The distance of $P(x, y)$ from the origin is

Beginner
A $x + y$ B $\sqrt{x + y}$ C $\sqrt{x^2 + y^2}$ D $x^2 + y^2$
Show answer & explanation

$\sqrt{x^2 + y^2}$

The origin is $(0, 0)$, so the differences are just x and y.

Q4. The distance between $(0, 0)$ and $(3, 4)$ is

Beginner
A 7 B $\sqrt7$ C 25 D 5
Show answer & explanation

5

$\sqrt{9 + 16} = 5$.

Q5. The distance between $(0, 0)$ and $(8, 15)$ is

Beginner
A 23 B 7 C 120 D 17
Show answer & explanation

17

$\sqrt{64 + 225} = \sqrt{289} = 17$.

Q6. The distance between $(2, 3)$ and $(2, 11)$ is

Beginner
A 8 B 14 C 6 D 0
Show answer & explanation

8

The x-coordinates match, so the distance is just the vertical gap.

Q7. The distance between $(1, 4)$ and $(7, 4)$ is

Beginner
A 8 B 6 C $\sqrt6$ D 11
Show answer & explanation

6

The y-coordinates match, so the distance is the horizontal gap.

Q8. Does the order of the two points change the distance?

Beginner
A yes, the sign changes B yes, only for negatives C no, because the differences are squared D only if one is the origin
Show answer & explanation

no, because the differences are squared

$(a - b)^2 = (b - a)^2$.

Q9. A distance can never be

Beginner
A a surd B negative C a whole number D zero
Show answer & explanation

negative

It is a length, so the positive root is always taken.

Q10. Simplified, $\sqrt{50}$ equals

Beginner
A $2\sqrt5$ B $5\sqrt2$ C $25\sqrt2$ D $\sqrt5 \times \sqrt{10}$
Show answer & explanation

$5\sqrt2$

$\sqrt{25 \times 2}$.

Q11. The distance between $(1, 1)$ and $(9, 7)$ is

Beginner
A 14 B 8 C 10 D 6
Show answer & explanation

10

$\sqrt{64 + 36} = \sqrt{100} = 10$.

Q12. Assertion (A): The distance between $(2, 5)$ and $(5, 9)$ is 5. Reason (R): The horizontal gap is 4 and the vertical gap is 3, and $\sqrt{4^2 + 3^2} = 7$.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is true but R is false

The gaps are 3 and 4 and the root is 5, not 7. The assertion is correct, the reason is not.

Q13. The squared distance between $(2, 1)$ and $(5, 5)$ is

Beginner
A 5 B 7 C 49 D 25
Show answer & explanation

25

$3^2 + 4^2 = 25$; the distance itself is 5.

Q14. Before squaring, the two differences between $(4, 9)$ and $(1, 5)$ are

Beginner
A 3 and 4 B −3 and −4 C 5 and 14 D −3 and 4
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−3 and −4

Taking the second point from the first: $1 - 4 = -3$ and $5 - 9 = -4$.

Q15. Two points have the same x-coordinate. The distance between them equals

Beginner
A the sum of their y-coordinates B the difference of their y-coordinates C zero D the difference of their x-coordinates
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the difference of their y-coordinates

The horizontal gap is zero, so only the vertical gap remains.

Q16. Find the distance between the points $(3, 1)$ and $(7, 4)$.

Beginner
Show answer & explanation

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Q17. Find the distance of the point $(-6, 8)$ from the origin.

Beginner
Show answer & explanation

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Q18. The section formula for the ratio $m : n$ from A to B gives the x-coordinate as

Beginner
A $\frac{mx_1 + nx_2}{m+n}$ B $\frac{mx_2 - nx_1}{m+n}$ C $\frac{mx_2 + nx_1}{m+n}$ D $\frac{x_1 + x_2}{m+n}$
Show answer & explanation

$\frac{mx_2 + nx_1}{m+n}$

The weights cross: m goes with the second point.

Q19. The midpoint of $(x_1, y_1)$ and $(x_2, y_2)$ is

Beginner
A $\left(\frac{x_1-x_2}{2}, \frac{y_1-y_2}{2}\right)$ B $(x_1+x_2, y_1+y_2)$ C $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ D $\left(\frac{x_1 y_1}{2}, \frac{x_2 y_2}{2}\right)$
Show answer & explanation

$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

The average of each coordinate.

Q20. The midpoint formula is the section formula with

Beginner
A $m = 2$, $n = 1$ B $m = 0$ C $m = n = 1$ D $n = 0$
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$m = n = 1$

Equal weights give an ordinary mean.

Q21. The midpoint of $(2, 4)$ and $(6, 8)$ is

Beginner
A $(8, 12)$ B $(4, 6)$ C $(2, 2)$ D $(3, 5)$
Show answer & explanation

$(4, 6)$

$\left(\frac{2+6}{2}, \frac{4+8}{2}\right)$.

Q22. The midpoint of $(-3, 5)$ and $(7, -1)$ is

Beginner
A $(2, 3)$ B $(5, 2)$ C $(2, 2)$ D $(-5, 3)$
Show answer & explanation

$(2, 2)$

$\left(\frac{-3+7}{2}, \frac{5-1}{2}\right)$.

Q23. In the section formula, what must you divide by?

Beginner
A $m + n$ B $m - n$ C $mn$ D 2 always
Show answer & explanation

$m + n$

The total number of parts in the ratio.

Q24. The two points of trisection divide a segment in the ratios

Beginner
A $1 : 3$ and $3 : 1$ B $1 : 2$ and $2 : 1$ C $1 : 1$ and $2 : 2$ D $1 : 2$ and $1 : 3$
Show answer & explanation

$1 : 2$ and $2 : 1$

Three equal parts means the first point is one part out of three.

Q25. A point on the x-axis has

Beginner
A $x = 0$ B $y = 0$ C $x = y$ D both coordinates zero
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$y = 0$

Height zero, so the y-coordinate vanishes.

Q26. To find where a segment meets the y-axis you set which expression to zero?

Beginner
A the y-coordinate B both coordinates C the ratio k D the x-coordinate
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the x-coordinate

Every point on the y-axis has x-coordinate zero.

Q27. An unknown ratio is best written as

Beginner
A $m : n$ B $1 : 1$ C $k : k$ D $k : 1$
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$k : 1$

It reduces two unknowns to one.

Q28. Is $3 : 2$ from A the same point as $2 : 3$ from A?

Beginner
A yes B no — they are different points C yes, if the segment is horizontal D only at the midpoint
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no — they are different points

The first is nearer B and the second nearer A.

Q29. Assertion (A): The midpoint formula does not need to be memorised separately. Reason (R): It is the section formula with the ratio $1 : 1$.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

Setting both weights to 1 turns the section formula into the average, exactly as R says.

Q30. The midpoint of $(0, 0)$ and $(10, 6)$ is

Beginner
A $(10, 6)$ B $(3, 5)$ C $(5, 3)$ D $(0, 0)$
Show answer & explanation

$(5, 3)$

Average each coordinate.

Q31. In the ratio $m : n$ from A to B, the part nearer B is

Beginner
A $n$ B $m$ C $m + n$ D $m - n$
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$n$

m measures the piece from A to the point, so n is the remaining piece to B.

Q32. Assertion (A): The midpoint of a segment is equidistant from its two endpoints. Reason (R): The midpoint divides the segment in the ratio $1 : 1$.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

Both A and R are true, and R is the correct explanation of A

Equal parts either side is exactly what a $1 : 1$ division means.

Q33. Find the midpoint of the line segment joining $(4, -3)$ and $(-2, 7)$.

Beginner
Show answer & explanation

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Q34. Three points are collinear when

Beginner
A all three distances are equal B the two shorter distances add to the longest C the squared distances add D they form a right angle
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the two shorter distances add to the longest

Equality leaves no room for a bend in the path.

Q35. To classify a quadrilateral from its vertices you compute

Beginner
A four sides only B two diagonals only C the perimeter D four sides and two diagonals
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four sides and two diagonals

The sides and the diagonals each decide part of the answer.

Q36. In a quadrilateral PQRS taken in order, the diagonals are

Beginner
A PR and QS B PQ and RS C QR and SP D PQ and QR
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PR and QS

Diagonals join opposite vertices — 1st to 3rd and 2nd to 4th.

Q37. A quadrilateral with all four sides equal and both diagonals equal is a

Beginner
A rhombus B rectangle C square D trapezium
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square

Equal sides give a rhombus; equal diagonals upgrade it to a square.

Q38. A quadrilateral with all four sides equal but unequal diagonals is a

Beginner
A square B rectangle C parallelogram only D rhombus
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rhombus

A squashed square keeps its side lengths.

Q39. A quadrilateral with opposite sides equal and equal diagonals is a

Beginner
A rectangle B square C rhombus D trapezium
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rectangle

Equal diagonals mean right angles at the corners.

Q40. Unequal diagonals distinguish a rectangle from a

Beginner
A square B rhombus C parallelogram D trapezium
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parallelogram

Both have opposite sides equal; only the rectangle has equal diagonals.

Q41. Which single check rules out the whole parallelogram family at once?

Beginner
A one side being longest B an odd perimeter C the diagonals not sharing a midpoint D a vertex at the origin
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the diagonals not sharing a midpoint

Bisecting diagonals are necessary for every parallelogram, rectangle, rhombus and square.

Q42. A point known to lie on the x-axis should be written as

Beginner
A $(0, y)$ B $(x, x)$ C $(0, 0)$ D $(x, 0)$
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$(x, 0)$

Writing it this way removes one unknown immediately.

Q43. For a question about a perimeter, the right tool is

Beginner
A the distance formula B the section formula C the midpoint formula D neither
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the distance formula

A perimeter is a sum of lengths.

Q44. For a question about trisection, the right tool is

Beginner
A the distance formula B Pythagoras C neither D the section formula
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the section formula

Trisection is a division in a ratio.

Q45. Assertion (A): Every square is also a rectangle. Reason (R): A rectangle requires opposite sides equal and equal diagonals, both of which a square satisfies.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

A square meets the rectangle's conditions and more, so the name still applies.

Q46. Which shape has four equal sides AND four right angles?

Beginner
A a rhombus B a square C a rectangle D a parallelogram
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a square

It satisfies both the rhombus and the rectangle conditions at once.

Q47. Assertion (A): To find a missing vertex of a parallelogram you equate two midpoints. Reason (R): The diagonals of a parallelogram bisect each other, so they share a midpoint.

Beginner
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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Both A and R are true, and R is the correct explanation of A

The shared midpoint is exactly what makes the two expressions equal.

Building up · 67 questions

Q48. The distance between $(-5, 7)$ and $(-1, 3)$ is

Intermediate
A $4\sqrt2$ B $4$ C $\sqrt8$ D $8$
Show answer & explanation

$4\sqrt2$

The gaps are 4 and $-4$, so $\sqrt{32} = 4\sqrt2$.

Q49. The distance between $(2, 3)$ and $(4, 1)$ is

Intermediate
A $\sqrt2$ B 4 C $2\sqrt2$ D $2$
Show answer & explanation

$2\sqrt2$

$\sqrt{4 + 4} = \sqrt8 = 2\sqrt2$.

Q50. How far is the point $(-3, -4)$ from the origin?

Intermediate
A 5 B 7 C 1 D 12
Show answer & explanation

5

$\sqrt{9 + 16} = 5$; the signs vanish on squaring.

Q51. Comparing $AB^2$ and $AC^2$ instead of $AB$ and $AC$ is valid because

Intermediate
A distances are positive, so equal squares mean equal lengths B squares are always larger C it is only an approximation D it works only for whole numbers
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distances are positive, so equal squares mean equal lengths

For non-negative numbers, $p^2 = q^2$ forces $p = q$.

Q52. To show three points form an isosceles triangle, you show that

Intermediate
A all three squared distances are equal B the largest square is the sum of the others C the points are collinear D two of the squared distances are equal
Show answer & explanation

two of the squared distances are equal

Two equal sides is the definition.

Q53. Which comparison proves a right angle without any square root?

Intermediate
A two squared distances are equal B the largest squared distance equals the sum of the other two C all three squared distances are equal D the squared distances add to zero
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the largest squared distance equals the sum of the other two

That is the converse of Pythagoras, applied to squares.

Q54. For $A(1, 7)$, $B(4, 2)$ and $C(-1, -1)$, the value of $AB^2$ is

Intermediate
A 34 B 68 C 25 D 17
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34

$3^2 + (-5)^2 = 9 + 25 = 34$.

Q55. The triangle with vertices $A(1, 7)$, $B(4, 2)$ and $C(-1, -1)$ is

Intermediate
A equilateral B isosceles and right-angled C scalene only D not a triangle
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isosceles and right-angled

$AB^2 = BC^2 = 34$ and $34 + 34 = 68 = CA^2$.

Q56. If the distance between $P(x, 2)$ and $Q(3, 6)$ is 5, then x is

Intermediate
A 0 only B 0 or 6 C 6 only D 3
Show answer & explanation

0 or 6

$(3-x)^2 = 9$ gives $3 - x = \pm3$.

Q57. A general point on the x-axis is written as

Intermediate
A $(0, y)$ B $(x, x)$ C $(0, 0)$ D $(x, 0)$
Show answer & explanation

$(x, 0)$

Every point on the x-axis has zero height.

Q58. Which coordinate is zero for every point on the y-axis?

Intermediate
A the y-coordinate B both C neither D the x-coordinate
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the x-coordinate

Points on the y-axis sit directly above or below the origin.

Q59. Assertion (A): The distance between $(4, 0)$ and $(0, 3)$ is 5. Reason (R): Both points lie on the axes, so the distance is the sum $4 + 3$.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is true but R is false

A is true — $\sqrt{16 + 9} = 5$ — but R is false, because the distance is the hypotenuse, not the sum of the legs.

Q60. Assertion (A): Two points with the same y-coordinate are separated by the difference of their x-coordinates. Reason (R): The vertical gap is then zero, so the formula reduces to $\sqrt{(x_2-x_1)^2} = |x_2 - x_1|$.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

Both A and R are true, and R is the correct explanation of A

With one gap zero the formula collapses to the other gap, exactly as R shows.

Q61. The squared distance between $(0, 5)$ and $(12, 0)$ is

Intermediate
A 13 B 17 C 144 D 169
Show answer & explanation

169

$144 + 25 = 169$.

Q62. For $A(0, 0)$, $B(4, 0)$ and $C(0, 3)$, the longest side is

Intermediate
A AB B BC C CA D all three are equal
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BC

$BC^2 = 25$, while $AB^2 = 16$ and $CA^2 = 9$.

Q63. If $PQ^2 = 45$, then PQ in simplest surd form is

Intermediate
A $5\sqrt3$ B $3\sqrt5$ C $9\sqrt5$ D $\sqrt{45}$ cannot be simplified
Show answer & explanation

$3\sqrt5$

$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt5$.

Q64. Find the distance between $(-4, 6)$ and $(2, -2)$.

Intermediate
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Q65. Find the distance between $(5, 2)$ and $(3, 4)$, leaving your answer as a surd.

Intermediate
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Q66. Show that the points $A(1, 7)$, $B(4, 2)$ and $C(-1, -1)$ are the vertices of an isosceles triangle.

Intermediate
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Q67. Find the value of x if the distance between $P(x, 2)$ and $Q(3, 6)$ is 5.

Intermediate
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Q68. Explain why comparing squared distances is usually better than comparing the distances themselves.

Intermediate
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Q69. Find the perimeter of the triangle with vertices $(0, 0)$, $(4, 0)$ and $(0, 3)$.

Intermediate
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Q70. (a) Derive the distance formula from Pythagoras' theorem, explaining each step. (b) Use it to find the distance between $(-3, 5)$ and $(2, -7)$. (c) State two properties of the formula that follow from the differences being squared.

Intermediate
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2 marks — (a) the construction, the right angle and Pythagoras applied. 1½ marks — (b) both gaps and d = 13. 1½ marks — (c) both properties.

Q71. The point dividing $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$ is

Intermediate
A $(2, 1)$ B $(1, 3)$ C $(3, 1)$ D $(1, -3)$
Show answer & explanation

$(1, 3)$

$x = \frac{8-3}{5} = 1$ and $y = \frac{-6+21}{5} = 3$.

Q72. The point dividing $(1, 2)$ and $(7, 8)$ in the ratio $1 : 2$ is

Intermediate
A $(3, 4)$ B $(4, 5)$ C $(5, 6)$ D $(2, 3)$
Show answer & explanation

$(3, 4)$

$x = \frac{7 + 2}{3} = 3$ and $y = \frac{8 + 4}{3} = 4$.

Q73. The first point of trisection of $(4, -1)$ and $(-2, -3)$ is

Intermediate
A $\left(0, -\frac73\right)$ B $\left(2, -\frac53\right)$ C $(1, -2)$ D $\left(2, -\frac73\right)$
Show answer & explanation

$\left(2, -\frac53\right)$

Ratio $1 : 2$ gives $x = \frac{-2 + 8}{3} = 2$ and $y = \frac{-3 - 2}{3} = -\frac53$.

Q74. The second point of trisection of $(4, -1)$ and $(-2, -3)$ is

Intermediate
A $\left(2, -\frac53\right)$ B $(0, -2)$ C $\left(-1, -\frac73\right)$ D $\left(0, -\frac73\right)$
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$\left(0, -\frac73\right)$

Ratio $2 : 1$ gives $x = \frac{-4 + 4}{3} = 0$ and $y = \frac{-6 - 1}{3} = -\frac73$.

Q75. If $P$ divides $A(2, 1)$ and $B(8, 7)$ in the ratio $1 : 2$, then P is

Intermediate
A $(6, 5)$ B $(5, 4)$ C $(3, 2)$ D $(4, 3)$
Show answer & explanation

$(4, 3)$

$x = \frac{8 + 4}{3} = 4$ and $y = \frac{7 + 2}{3} = 3$.

Q76. Setting the y-coordinate of the section formula to zero finds where a segment

Intermediate
A crosses the x-axis B crosses the y-axis C meets the origin D is bisected
Show answer & explanation

crosses the x-axis

On the x-axis every point has $y = 0$.

Q77. The x-axis divides the join of $(1, -5)$ and $(-4, 5)$ in the ratio

Intermediate
A $1 : 2$ B $2 : 1$ C $1 : 1$ D $5 : 4$
Show answer & explanation

$1 : 1$

$\frac{5k - 5}{k+1} = 0$ gives $k = 1$.

Q78. The point where the x-axis cuts the join of $(1, -5)$ and $(-4, 5)$ is

Intermediate
A $\left(-\frac32, 0\right)$ B $\left(\frac32, 0\right)$ C $(0, 0)$ D $\left(0, -\frac32\right)$
Show answer & explanation

$\left(-\frac32, 0\right)$

The ratio is $1 : 1$, so it is the midpoint $\left(\frac{1-4}{2}, 0\right)$.

Q79. If the midpoint of $(a, 4)$ and $(6, b)$ is $(5, 7)$, then a and b are

Intermediate
A $a = 5$, $b = 7$ B $a = 4$, $b = 10$ C $a = 6$, $b = 4$ D $a = 10$, $b = 4$
Show answer & explanation

$a = 4$, $b = 10$

$\frac{a+6}{2} = 5$ gives $a = 4$, and $\frac{4+b}{2} = 7$ gives $b = 10$.

Q80. Why is an unknown ratio written as $k : 1$?

Intermediate
A it is always true B it makes the point the midpoint C it avoids fractions D it reduces two unknowns to one
Show answer & explanation

it reduces two unknowns to one

Any ratio $m : n$ equals $\frac mn : 1$.

Q81. After finding a ratio from the x-coordinates, you should

Intermediate
A verify it using the y-coordinates B stop, since one coordinate is enough C double the ratio D take its reciprocal
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verify it using the y-coordinates

The check costs one line and catches nearly every slip.

Q82. Assertion (A): The points of trisection divide a segment in the ratios $1 : 3$ and $3 : 1$. Reason (R): Trisection cuts a segment into three equal parts.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is false but R is true

R is true, but the ratios are $1 : 2$ and $2 : 1$ — one part against the remaining two — so A is false.

Q83. Assertion (A): To find where a segment meets the y-axis, set the section formula's y-coordinate to zero. Reason (R): Every point on the y-axis has x-coordinate zero.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
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A is false but R is true

It is the x-coordinate that vanishes on the y-axis, so the assertion is false. The reason states the property correctly.

Q84. The point dividing $(1, 1)$ and $(5, 9)$ in the ratio $3 : 1$ is

Intermediate
A $(2, 3)$ B $(3, 5)$ C $(4, 5)$ D $(4, 7)$
Show answer & explanation

$(4, 7)$

$x = \frac{15 + 1}{4} = 4$ and $y = \frac{27 + 1}{4} = 7$.

Q85. A point that is one-quarter of the way from A to B divides AB in the ratio

Intermediate
A $1 : 4$ B $3 : 1$ C $1 : 3$ D $1 : 2$
Show answer & explanation

$1 : 3$

One part travelled and three remaining.

Q86. If the midpoint of $(2, p)$ and $(q, 8)$ is $(5, 6)$, then p and q are

Intermediate
A $p = 8$, $q = 4$ B $p = 4$, $q = 8$ C $p = 6$, $q = 5$ D $p = 5$, $q = 6$
Show answer & explanation

$p = 4$, $q = 8$

$\frac{p+8}{2} = 6$ gives $p = 4$, and $\frac{2+q}{2} = 5$ gives $q = 8$.

Q87. Locate the point that splits the segment from $(-1, 7)$ to $(4, -3)$ in the ratio $2 : 3$, and give its coordinates.

Intermediate
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Q88. Explain why the section formula has the weights crossed over.

Intermediate
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Q89. If the midpoint of the segment joining $(a, 4)$ and $(6, b)$ is $(5, 7)$, find a and b.

Intermediate
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Q90. Find the point which divides the join of $(1, 2)$ and $(7, 10)$ in the ratio $3 : 1$.

Intermediate
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Q91. Why is it wrong to use the ratio $1 : 3$ for a point of trisection?

Intermediate
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Q92. (a) State the section formula and explain what each letter means. (b) Derive the midpoint formula from it. (c) Find the midpoint and the two trisection points of the segment joining $A(2, 1)$ and $B(8, 7)$.

Intermediate
Show answer & explanation

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1½ marks — (a) the formula with every letter explained. 1 mark — (b) the substitution $m = n = 1$. 2½ marks — (c) the midpoint and both trisection points.

Q93. For collinear points with $AB^2 = 10$, $BC^2 = 40$ and $AC^2 = 90$, the sum $10 + 40$ is

Intermediate
A equal to 90 B not equal to 90, which is why squares must not be used C equal to $AC$ D irrelevant to the points
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not equal to 90, which is why squares must not be used

Squaring does not respect addition, so collinearity needs the distances themselves.

Q94. The points $(-2, 3)$, $(1, 2)$ and $(7, 0)$ are

Intermediate
A not collinear B the vertices of a right triangle C collinear D the vertices of an equilateral triangle
Show answer & explanation

collinear

$\sqrt{10} + 2\sqrt{10} = 3\sqrt{10}$.

Q95. The quadrilateral with vertices $(1,1)$, $(5,1)$, $(5,5)$, $(1,5)$ is a

Intermediate
A rhombus B rectangle only C square D parallelogram only
Show answer & explanation

square

All four sides are 16 and both diagonals are 32.

Q96. The quadrilateral with vertices $(0,0)$, $(4,3)$, $(0,6)$, $(-4,3)$ is a

Intermediate
A square B rhombus C rectangle D trapezium
Show answer & explanation

rhombus

Sides all 25, diagonals 36 and 64 — equal sides but unequal diagonals.

Q97. The quadrilateral with vertices $(0,0)$, $(6,0)$, $(6,3)$, $(0,3)$ is a

Intermediate
A rectangle B square C rhombus D parallelogram only
Show answer & explanation

rectangle

Opposite sides equal (36 and 9) and both diagonals 45.

Q98. Three vertices of a parallelogram in order are $(1,2)$, $(4,3)$ and $(6,6)$. The fourth is

Intermediate
A $(9, 7)$ B $(5, 3)$ C $(3, 5)$ D $(2, 1)$
Show answer & explanation

$(3, 5)$

Equating the midpoints of AC and BD gives $4 + x = 7$ and $3 + y = 8$.

Q99. Why must the vertices be taken in order?

Intermediate
A otherwise the distances change B otherwise the 'diagonals' would actually be sides C order never matters D to keep the perimeter positive
Show answer & explanation

otherwise the 'diagonals' would actually be sides

The order is what identifies which vertices are opposite.

Q100. If a shape satisfies the conditions for both a rectangle and a rhombus, it is a

Intermediate
A parallelogram B trapezium C kite D square
Show answer & explanation

square

Equal sides and equal diagonals together define a square.

Q101. To find a point equidistant from two given points, you

Intermediate
A add the two distances B take the midpoint of their squares C use the section formula D set the two squared distances equal
Show answer & explanation

set the two squared distances equal

The $x^2$ terms then cancel and a linear equation remains.

Q102. A question asks what kind of triangle three points form. The right tool is

Intermediate
A the distance formula B the section formula C the midpoint formula D the ratio $k : 1$
Show answer & explanation

the distance formula

Classifying a triangle compares side lengths.

Q103. A question asks for the fourth vertex of a parallelogram. The right tool is

Intermediate
A the distance formula B the collinearity test C Pythagoras D the section formula, via equal midpoints
Show answer & explanation

the section formula, via equal midpoints

The diagonals bisect each other.

Q104. Assertion (A): Four equal sides prove a quadrilateral is a square. Reason (R): A rhombus also has four equal sides.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is false but R is true

R is true and is exactly the counterexample that makes A false — the diagonals must be checked too.

Q105. Assertion (A): Collinearity can be tested using squared distances. Reason (R): Comparing squared distances is valid whenever lengths are being compared for equality.

Intermediate
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

A is false but R is true

R is true for EQUALITY, but collinearity needs a SUM, and squaring does not respect addition — so A is false.

Q106. The quadrilateral $(4,0)$, $(8,0)$, $(8,4)$, $(4,4)$ is a

Intermediate
A rhombus only B rectangle only C square D parallelogram only
Show answer & explanation

square

All four squared sides are 16 and both squared diagonals are 32.

Q107. Three collinear points enclose a region of area

Intermediate
A one square unit B zero C half the longest side D any positive value
Show answer & explanation

zero

Points on a single line cannot enclose anything.

Q108. If $AB = 3\sqrt2$, $BC = 2\sqrt2$ and $AC = 5\sqrt2$, the points are

Intermediate
A not collinear B collinear, with B between A and C C the vertices of a right triangle D coincident
Show answer & explanation

collinear, with B between A and C

$3\sqrt2 + 2\sqrt2 = 5\sqrt2$, so the two shorter lengths add to the longest.

Q109. Writing each distance as a multiple of the same surd helps because

Intermediate
A it changes the lengths B it avoids the distance formula C the addition becomes obvious and stays exact D decimals are more accurate
Show answer & explanation

the addition becomes obvious and stays exact

Comparing $\sqrt{10} + 2\sqrt{10}$ with $3\sqrt{10}$ needs no approximation at all.

Q110. Show that the points $A(-2, 3)$, $B(1, 2)$ and $C(7, 0)$ are collinear.

Intermediate
Show answer & explanation

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Q111. Show that $(1,1)$, $(5,1)$, $(5,5)$ and $(1,5)$ are the vertices of a square.

Intermediate
Show answer & explanation

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Q112. Explain why four equal sides do not prove that a quadrilateral is a square.

Intermediate
Show answer & explanation

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Q113. State which formula you would use for each: (i) the perimeter of a triangle, (ii) the midpoint of a segment, (iii) whether three points are collinear.

Intermediate
Show answer & explanation

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Q114. Why does a question say 'taken in order' when listing the vertices of a quadrilateral?

Intermediate
Show answer & explanation

—

Topper level · 57 questions

Q115. The point on the x-axis equidistant from $(2, -5)$ and $(-2, 9)$ is

Advanced
A $(7, 0)$ B $(0, -7)$ C $(-7, 0)$ D $(-7, 2)$
Show answer & explanation

$(-7, 0)$

Setting $PA^2 = PB^2$ with $P(x, 0)$ gives $29 - 4x = 85 + 4x$, so $x = -7$.

Q116. Why does setting two squared distances equal give a LINEAR equation?

Advanced
A squares are always linear B the $x^2$ terms on both sides cancel C the distances are equal D the points lie on a line
Show answer & explanation

the $x^2$ terms on both sides cancel

Expanding both sides produces the same $x^2$, which subtracts out.

Q117. The points $(1, 5)$, $(2, 3)$ and $(-2, -11)$ are

Advanced
A collinear B the vertices of an equilateral triangle C not collinear D the vertices of a right triangle
Show answer & explanation

not collinear

$AB = \sqrt5$ and $BC = \sqrt{212}$, while $AC = \sqrt{265}$; since $\sqrt5 + \sqrt{212} \ne \sqrt{265}$, they are not collinear.

Q118. If $(a, 0)$ is equidistant from $(3, 4)$ and $(-1, 2)$, then a is

Advanced
A 2 B 3 C 1 D 2.5
Show answer & explanation

2.5

$(3-a)^2 + 16 = (-1-a)^2 + 4$ gives $25 - 6a = 5 + 2a$, so $8a = 20$.

Q119. A point equidistant from all three vertices of a triangle is

Advanced
A the centroid B the midpoint of a side C the circumcentre D always the origin
Show answer & explanation

the circumcentre

Equal distances to all three vertices is the defining property of the circumcentre.

Q120. The triangle with vertices $(0, 0)$, $(4, 0)$ and $(0, 3)$ has perimeter

Advanced
A 10 B 14 C 12 D $7 + \sqrt5$
Show answer & explanation

12

The sides are 4, 3 and $\sqrt{16+9} = 5$, so $4 + 3 + 5 = 12$.

Q121. If the distance between $(2, -3)$ and $(10, y)$ is 10, then y is

Advanced
A 3 only B −9 only C 6 D 3 or −9
Show answer & explanation

3 or −9

$64 + (y+3)^2 = 100$ gives $(y+3)^2 = 36$, so $y + 3 = \pm6$.

Q122. The points $(3, 0)$, $(6, 4)$ and $(-1, 3)$ form

Advanced
A an equilateral triangle B a scalene triangle with no right angle C a straight line D an isosceles right triangle
Show answer & explanation

an isosceles right triangle

The squared sides are 25, 50 and 25, and $25 + 25 = 50$.

Q123. Assertion (A): Three points can be tested for a right angle without finding any square root. Reason (R): The converse of Pythagoras can be applied to the squared distances directly.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

Both A and R are true, and R is the correct explanation of A

Comparing squares is exactly the test, and it never needs a root.

Q124. Any point at distance 13 from the origin satisfies

Advanced
A $x^2 + y^2 = 169$ B $x + y = 13$ C $x^2 + y^2 = 13$ D $x^2 - y^2 = 169$
Show answer & explanation

$x^2 + y^2 = 169$

Squaring $\sqrt{x^2+y^2} = 13$ gives the equation of that circle.

Q125. Assertion (A): The distance between $(a, b)$ and $(b, a)$ is zero when $a = b$. Reason (R): The distance between two points is zero exactly when the points coincide.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

Both A and R are true, and R is the correct explanation of A

If $a = b$ the two points are the same, and R states the general rule behind that.

Q126. Show that the triangle in the previous question is also right-angled, and name the right angle.

Advanced
Show answer & explanation

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Q127. Which point of the x-axis stands the same distance from $(2, -5)$ as it does from $(-2, 9)$?

Advanced
Show answer & explanation

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Q128. Find the values of y for which the distance between $(2, -3)$ and $(10, y)$ is 10.

Advanced
Show answer & explanation

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Q129. The points $A(3, 0)$, $B(6, 4)$ and $C(-1, 3)$ are plotted. (a) Find $AB^2$, $BC^2$ and $CA^2$. (b) Determine what kind of triangle ABC is, justifying every claim. (c) Find the perimeter of the triangle, leaving surds in exact form.

Advanced
Show answer & explanation

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1½ marks — (a) all three squared distances. 2 marks — (b) isosceles and right-angled, each justified, with the right angle located. 1½ marks — (c) the exact perimeter.

Q130. (a) Find the point on the y-axis which is equidistant from $A(6, 5)$ and $B(-4, 3)$. (b) Explain why setting the two squared distances equal leaves a linear equation. (c) Verify your answer by computing both distances.

Advanced
Show answer & explanation

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2 marks — (a) the setup and y = 9. 1½ marks — (b) the cancellation explained. 1½ marks — (c) both squared distances computed and shown equal.

Q131. A surveyor plots three markers on a field at $P(0, 0)$, $Q(8, 0)$ and $R(0, 6)$, with all coordinates in metres. (a) Find the distances PQ, PR and QR. (b) Show that the markers form a right-angled triangle and state where the right angle is. (c) A fourth marker S is to be placed so that PQSR is a rectangle. Give its coordinates, and find the distance from P to S.

Advanced
Show answer & explanation

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1½ marks — (a) all three distances. 1½ marks — (b) the Pythagoras check and the location of the right angle. 2 marks — (c) S = (8, 6) and PS = 10 with the diagonal comment.

Q132. The point $(-1, 6)$ divides the join of $(-3, 10)$ and $(6, -8)$ in the ratio

Advanced
A $2 : 7$ B $7 : 2$ C $1 : 3$ D $3 : 1$
Show answer & explanation

$2 : 7$

With ratio $k : 1$: $\frac{6k - 3}{k + 1} = -1$ gives $7k = 2$.

Q133. Taking $(1, 2)$, $(4, y)$, $(x, 6)$ and $(3, 5)$ as the corners of a parallelogram, named in order, x and y come to

Advanced
A $x = 3$, $y = 6$ B $x = 5$, $y = 4$ C $x = 6$, $y = 3$ D $x = 4$, $y = 5$
Show answer & explanation

$x = 6$, $y = 3$

The diagonals of a parallelogram bisect each other, so the midpoints of the two diagonals coincide.

Q134. The diagonals of a parallelogram

Advanced
A are equal B bisect each other C are perpendicular D are parallel
Show answer & explanation

bisect each other

That is what makes the midpoint method work for parallelogram problems.

Q135. If $A(1, 2)$ and $B(7, 10)$, the point three-quarters of the way from A to B divides AB in the ratio

Advanced
A $1 : 3$ B $3 : 4$ C $4 : 3$ D $3 : 1$
Show answer & explanation

$3 : 1$

Three parts travelled and one remaining gives $3 : 1$.

Q136. The point three-quarters of the way from $A(1, 2)$ to $B(7, 10)$ is

Advanced
A $(4, 6)$ B $\left(\frac{11}{2}, 8\right)$ C $\left(\frac52, 4\right)$ D $(6, 9)$
Show answer & explanation

$\left(\frac{11}{2}, 8\right)$

Ratio $3 : 1$ gives $x = \frac{21 + 1}{4} = \frac{11}{2}$ and $y = \frac{30 + 2}{4} = 8$.

Q137. A negative value of k when solving for a ratio means

Advanced
A the point is the midpoint B the calculation is always wrong C the point lies outside the segment D the segment is vertical
Show answer & explanation

the point lies outside the segment

It indicates external division, on the extension of the line.

Q138. The y-axis divides the join of $(-4, -6)$ and $(2, 3)$ in the ratio

Advanced
A $2 : 1$ B $1 : 2$ C $3 : 2$ D $2 : 3$
Show answer & explanation

$2 : 1$

$\frac{2k - 4}{k+1} = 0$ gives $2k = 4$, so $k = 2$.

Q139. If the midpoint of the join of $(2a, 4)$ and $(-2, 3b)$ is $(1, 5)$, then a and b are

Advanced
A $a = 1$, $b = 3$ B $a = 2$, $b = 3$ C $a = 2$, $b = 2$ D $a = 3$, $b = 2$
Show answer & explanation

$a = 2$, $b = 2$

$\frac{2a - 2}{2} = 1$ gives $a = 2$, and $\frac{4 + 3b}{2} = 5$ gives $b = 2$.

Q140. Assertion (A): Finding a ratio needs only one of the two coordinate equations. Reason (R): The other coordinate provides an independent check on the answer.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

Both A and R are true, but R is not the correct explanation of A

Both statements are true — one equation does determine k — but R describes a check rather than explaining why one equation suffices.

Q141. The centroid of a triangle divides each median in the ratio

Advanced
A $1 : 2$ from the vertex B $1 : 1$ C $3 : 1$ D $2 : 1$ from the vertex
Show answer & explanation

$2 : 1$ from the vertex

It sits two-thirds of the way along each median from the vertex.

Q142. The section formula gives a point BETWEEN A and B when

Advanced
A m is negative B n is negative C m equals n D both m and n are positive
Show answer & explanation

both m and n are positive

Internal division needs both parts of the ratio positive.

Q143. Split the segment from $(4, -1)$ to $(-2, -3)$ into three equal parts, and name the two points where the cuts fall.

Advanced
Show answer & explanation

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Q144. In what ratio does the point $(-1, 6)$ divide the join of $(-3, 10)$ and $(6, -8)$?

Advanced
Show answer & explanation

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Q145. In what ratio does the x-axis divide the line segment joining $A(1, -5)$ and $B(-4, 5)$? Find the point of division.

Advanced
Show answer & explanation

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Q146. The points $(1, 2)$, $(4, y)$, $(x, 6)$ and $(3, 5)$ are the corners of a parallelogram, named in order. Work out x and y.

Advanced
Show answer & explanation

—

Q147. In what ratio does the y-axis divide the join of $(-4, -6)$ and $(2, 3)$?

Advanced
Show answer & explanation

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Q148. (a) Explain the advantage of writing an unknown ratio as $k : 1$. (b) Find the ratio in which $P(-1, 6)$ divides the join of $A(-3, 10)$ and $B(6, -8)$. (c) Verify your answer using the other coordinate.

Advanced
Show answer & explanation

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1½ marks — (a) the reason one unknown suffices. 2 marks — (b) the equation and $k = \frac27$. 1½ marks — (c) the verification.

Q149. (a) Explain how to find the ratio in which the x-axis divides a segment. (b) Find the ratio in which the x-axis divides the join of $A(1, -5)$ and $B(-4, 5)$, and the point of division. (c) Find the ratio in which the y-axis divides the join of $C(-4, -6)$ and $D(2, 3)$.

Advanced
Show answer & explanation

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1½ marks — (a) the method with the y = 0 reason. 2 marks — (b) the ratio and the point. 1½ marks — (c) the ratio for the y-axis.

Q150. During Sports Day on a rectangular ground, flower pots are placed 1 m apart along one edge. Niharika posts a green flag at $(2, 25)$ and Preet posts a red flag at $(8, 20)$, distances in metres. (a) Find the distance between the two flags. (b) Rashmi must post a blue flag exactly halfway between them. Where should she post it? (c) A fourth flag is to go one-third of the way from the green flag to the red flag. Find its position.

Advanced
Show answer & explanation

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1½ marks — (a) the distance in surd form. 1½ marks — (b) the midpoint. 2 marks — (c) recognising the ratio 1 : 2 and both coordinates.

Q151. The points $(0, 1)$, $(2, 4)$ and $(5, 7)$ are

Advanced
A not collinear B collinear C the vertices of a square D coincident
Show answer & explanation

not collinear

$AB = \sqrt{13}$ and $BC = \sqrt{18}$, but $AC = \sqrt{61}$, and $\sqrt{13} + \sqrt{18} e \sqrt{61}$.

Q152. The quadrilateral $(0,0)$, $(5,0)$, $(7,3)$, $(2,3)$ is a

Advanced
A rectangle B rhombus C square D parallelogram
Show answer & explanation

parallelogram

Opposite sides are equal (25 and 13) but the diagonals are 58 and 18 — unequal.

Q153. Three vertices of a rectangle in order are $(0,0)$, $(8,0)$ and $(8,6)$. The fourth is

Advanced
A $(6, 0)$ B $(8, 8)$ C $(0, 6)$ D $(6, 8)$
Show answer & explanation

$(0, 6)$

Equating the midpoints of the diagonals gives $(0, 6)$.

Q154. A point on the y-axis equidistant from $(6, 5)$ and $(-4, 3)$ is

Advanced
A $(0, 4)$ B $(0, 9)$ C $(9, 0)$ D $(0, -9)$
Show answer & explanation

$(0, 9)$

$36 + (5-y)^2 = 16 + (3-y)^2$ gives $61 - 10y = 25 - 6y$, so $y = 9$.

Q155. For the quadrilateral $(1,1),(5,1),(5,5),(1,5)$, the squared diagonals are

Advanced
A 32 and 16 B both 16 C 36 and 32 D both 32
Show answer & explanation

both 32

$(1,1)$ to $(5,5)$ gives $16 + 16 = 32$, and $(5,1)$ to $(1,5)$ gives the same.

Q156. If three points are collinear, the triangle they 'form' has

Advanced
A a right angle B zero area, so it is not a triangle C equal sides D a positive area
Show answer & explanation

zero area, so it is not a triangle

Collinear points cannot enclose a region.

Q157. Which pair of checks distinguishes a rectangle from a general parallelogram?

Advanced
A the four sides B the perimeter C the two diagonals D the midpoints
Show answer & explanation

the two diagonals

Both have opposite sides equal; only a rectangle has equal diagonals.

Q158. The fourth vertex D of a parallelogram ABCD with $A(-2,1)$, $B(1,0)$ and $C(4,3)$ is

Advanced
A $(7, 2)$ B $(1, 4)$ C $(1, -4)$ D $(-5, 4)$
Show answer & explanation

$(1, 4)$

Midpoint of AC is $(1, 2)$; setting the midpoint of BD equal gives $1 + x = 2$ and $0 + y = 4$.

Q159. Assertion (A): A classification answer should give the most specific shape name. Reason (R): A square also satisfies the definitions of a rectangle, a rhombus and a parallelogram.

Advanced
A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true
Show answer & explanation

Both A and R are true, and R is the correct explanation of A

Because looser names are also true, only the most specific one distinguishes the shape.

Q160. A quadrilateral whose diagonals do NOT have the same midpoint is

Advanced
A not a parallelogram at all B a rhombus C a rectangle D a square
Show answer & explanation

not a parallelogram at all

Bisecting diagonals are necessary for a parallelogram, so the test fails at the first step.

Q161. The point on the x-axis equidistant from $(7, 6)$ and $(-3, 4)$ is

Advanced
A $(0, 3)$ B $(-3, 0)$ C $(3, 0)$ D $(2, 0)$
Show answer & explanation

$(3, 0)$

$85 - 14x = 25 + 6x$ gives $20x = 60$.

Q162. Explain why collinearity must not be tested using squared distances.

Advanced
Show answer & explanation

—

Q163. Do $(1, 5)$, $(2, 3)$ and $(-2, -11)$ all lie along one straight line? Justify your answer.

Advanced
Show answer & explanation

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Q164. Show that $(0,0)$, $(4,3)$, $(0,6)$ and $(-4,3)$ form a rhombus but not a square.

Advanced
Show answer & explanation

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Q165. Three vertices of a parallelogram taken in order are $(1, 2)$, $(4, 3)$ and $(6, 6)$. Find the fourth vertex.

Advanced
Show answer & explanation

—

Q166. Find the point on the y-axis which is equidistant from $(6, 5)$ and $(-4, 3)$.

Advanced
Show answer & explanation

—

Q167. Classify the quadrilateral with vertices $(0,0)$, $(5,0)$, $(7,3)$ and $(2,3)$, justifying your answer.

Advanced
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Q168. (a) State the test for three points to be collinear, and explain why squared distances must not be used. (b) Determine whether $A(3, 1)$, $B(6, 4)$ and $C(8, 6)$ are collinear. (c) Determine whether $P(1, 5)$, $Q(2, 3)$ and $R(-2, -11)$ are collinear.

Advanced
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1½ marks — (a) the test and the reason squares fail. 2 marks — (b) all three distances and the conclusion. 1½ marks — (c) the distances and the conclusion.

Q169. (a) List the four classifications of a quadrilateral in terms of its sides and diagonals. (b) Classify $(1,1)$, $(5,1)$, $(5,5)$, $(1,5)$, showing all six squared lengths. (c) Classify $(0,0)$, $(4,3)$, $(0,6)$, $(-4,3)$, and explain what distinguishes it from your answer to (b).

Advanced
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1½ marks — (a) all four classifications. 2 marks — (b) six lengths and the name. 1½ marks — (c) the name and the distinguishing comparison.

Q170. (a) State the property of a parallelogram used to find a missing vertex. (b) Three vertices of parallelogram ABCD taken in order are $A(-2, 1)$, $B(1, 0)$ and $C(4, 3)$. Find D. (c) Verify your answer by checking that the opposite sides are equal.

Advanced
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1 mark — (a) the bisecting-diagonals property. 2 marks — (b) both midpoints equated and D found. 2 marks — (c) both pairs of opposite sides checked.

Q171. A town planner marks four corners of a proposed park at $P(0, 0)$, $Q(6, 0)$, $R(6, 3)$ and $S(0, 3)$, in metres, taken in order. (a) Find the four squared side lengths and the two squared diagonals. (b) Classify the park, justifying your answer fully. (c) Find the point where the two diagonals cross, and explain how you know they cross there.

Advanced
Show answer & explanation

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1½ marks — (a) all six squared lengths. 2 marks — (b) the classification with both comparisons stated. 1½ marks — (c) the crossing point and the bisecting reason.

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