CBSE Maths › Class 10 PYQs › Coordinate Geometry

Coordinate Geometry: 3 marks Questions (CBSE Class 10)

44 different 3 marks questions on Coordinate Geometry from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (97)2 marks (57)3 marks (44)4 marks (11)

Q313 marksShort AnswerCoordinate Geometry CBSE 2026 · Basic 430/4/1

The vertices of a rhombus ABCD are A(– 3, – 4), B(5, – 3), C(1, 4) and D(– 7, 3). Find the length of both the diagonals. Hence, find area of the rhombus ABCD.

Show answer & solution

Answer: AC =

45

units, BD =

65

units, area = 60 sq. units

AC = $(1 + 3)^{2} + (4 + 4)^{2} = 16 + 64 = 80 = 45$ units.
BD = $(- 7 - 5)^{2} + (3 + 3)^{2} = 144 + 36 = 180 = 65$ units.
Area = $\frac{1}{2} \times A C \times B D = \frac{1}{2} \times 45 \times 65 = 60$ sq. units.

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q31 (OR) (OR)3 marksShort AnswerCoordinate Geometry CBSE 2026 · Basic 430/4/1

The line segment joining the points A (– 5, 1) and B (7, 6) is trisected at the points P and Q such that P is nearer to A. If P lies on the line $x$ + y = k, then find the value of k.

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Answer: k =

\frac{5}{3}

P divides AB in the ratio 1 : 2.
P = $(\frac{1 ( 7 ) + 2 ( - 5 )}{3}, \frac{1 ( 6 ) + 2 ( 1 )}{3}) = (- 1, \frac{8}{3})$ .
P lies on $x + y = k$ : $k = - 1 + \frac{8}{3} = \frac{5}{3}$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q273 marksShort AnswerCoordinate Geometry CBSE 2026 · Basic 430/5/1

Points P(6, 0), Q(2, 8) and R(−2, 4) are vertices of $△ P QR$ . It is given that MN $∥$ QR such that $\frac{P M}{M Q} = \frac{1}{3}$ . Using distance formula and ratio formula, show that $\frac{M N}{QR} = \frac{1}{4}$ .

Diagram for CBSE 2026 Class 10 Maths question 27

Show answer & solution

Answer: Proved: M(5, 2), N(4, 1), MN

= 2

, QR

= 42

, so

\frac{M N}{QR} = \frac{1}{4}

M divides PQ in 1 : 3: $M = (\frac{1 \times 2 + 3 \times 6}{4}, \frac{1 \times 8 + 3 \times 0}{4}) = (5, 2)$
Since MN $∥$ QR, by BPT $\frac{P N}{N R} = \frac{P M}{M Q} = \frac{1}{3}$ .
$N = (\frac{1 \times ( - 2 ) + 3 \times 6}{4}, \frac{1 \times 4 + 3 \times 0}{4}) = (4, 1)$
MN $= (5 - 4)^{2} + (2 - 1)^{2} = 2$ ; QR $= (2 + 2)^{2} + (8 - 4)^{2} = 42$
$\frac{M N}{QR} = \frac{2}{4 2} = \frac{1}{4}$

Also asked in: 2026 Basic 430/5/2

Q293 marksShort AnswerCoordinate Geometry CBSE 2026 · Basic 430/5/3

Determine the ratio in which the line $2 x + y = 6$ divides the line segment joining the points (1, 3) and (2, 5).

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Answer: 1 : 3 (internally)

Let the line divide the segment in the ratio k : 1. The point is $(\frac{2 k + 1}{k + 1}, \frac{5 k + 3}{k + 1})$ .
It lies on $2 x + y = 6$ : $\frac{2 ( 2 k + 1 ) + 5 k + 3}{k + 1} = 6$
$9 k + 5 = 6 k + 6 \Rightarrow k = \frac{1}{3}$
Ratio 1 : 3 (the point is $(\frac{5}{4}, \frac{7}{2})$ ).

Q273 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/1/1

Find the ratio in which the $x$ -axis divides the line segment joining the points $(- 6, 5)$ and $(- 4, - 1)$ . Also, find the point of intersection.

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Answer: 5 : 1;

(- \frac{13}{3}, 0)

Let the $x$ -axis divide the segment in the ratio $k : 1$ .
The point is $(\frac{- 4 k - 6}{k + 1}, \frac{- k + 5}{k + 1})$ .
On the $x$ -axis, $y = 0$ : $- k + 5 = 0$ , so $k = 5$ . Ratio = 5 : 1.
$x = \frac{- 20 - 6}{6} = - \frac{26}{6} = - \frac{13}{3}$ .
Point of intersection $= (- \frac{13}{3}, 0)$ .

Also asked in: 2026 Standard 30/1/3

Q313 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/1/2

Parthi and Alisha found a treasure that is exactly on the straight line joining their locations. Parthi's location is at point $(- 6, - 5)$ and Alisha's location is at point $(10, 11)$ . The distance from the treasure to Parthi's location is three times that of the distance to Alisha's location. Find the coordinates of the location of the treasure.

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Answer:

(6, 7)

Let P $(- 6, - 5)$ , A $(10, 11)$ and treasure T on PA with $T P = 3 T A$ .
So T divides PA internally in the ratio $P T : T A = 3 : 1$ .
$x = \frac{3 \times 10 + 1 \times ( - 6 )}{3 + 1} = \frac{24}{4} = 6$ .
$y = \frac{3 \times 11 + 1 \times ( - 5 )}{3 + 1} = \frac{28}{4} = 7$ .
The treasure is at $(6, 7)$ .

Q273 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/2/1

Find the coordinates of the points of trisection of the line segment joining the points A(-1, 4) and B(-3, -2).

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Answer:

(- \frac{5}{3}, 2)

and

(- \frac{7}{3}, 0)

Let P and Q divide AB in the ratios 1 : 2 and 2 : 1.
$P = (\frac{1 ( - 3 ) + 2 ( - 1 )}{3}, \frac{1 ( - 2 ) + 2 ( 4 )}{3}) = (- \frac{5}{3}, 2)$
$Q = (\frac{2 ( - 3 ) + 1 ( - 1 )}{3}, \frac{2 ( - 2 ) + 1 ( 4 )}{3}) = (- \frac{7}{3}, 0)$

Also asked in: 2026 Standard 30/2/2

Q293 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/2/3

The three vertices of a rhombus PQRS are P(2, -3), Q(6, 5) and R(-2, 1). Find the coordinates of the fourth vertex S and coordinates of the point where both the diagonals PR and QS intersect.

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Answer: S(-6, -7); diagonals intersect at (0, -1)

Diagonals of a rhombus bisect each other.
Mid-point of PR $= (\frac{2 - 2}{2}, \frac{- 3 + 1}{2}) = (0, - 1)$ ; this is the point of intersection.
Let S = (x, y). Mid-point of QS = (0, -1): $\frac{6 + x}{2} = 0$ , $\frac{5 + y}{2} = - 1$ .
x = -6, y = -7, so S = (-6, -7).

Q303 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/3/1

A point P(x, 7) divides a line segment joining the points A(– 5, 4) and B(7, 9) in a certain ratio. Find the ratio and hence find the value of x.

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Answer: Ratio 3 : 2;

x = \frac{11}{5}

Let P divide AB in the ratio $k : 1$
y-coordinate: $\frac{9 k + 4}{k + 1} = 7$ , so $9 k + 4 = 7 k + 7$ , $k = \frac{3}{2}$
Ratio $= 3 : 2$
$x = \frac{3 \times 7 + 2 \times ( - 5 )}{3 + 2} = \frac{11}{5}$

Q273 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/3/2

A point P divides the line segment joining the points A(– 3, 5) and B(7, – 4) in a certain ratio. If the point P lies on the line y = 2x, then find the ratio AP : PB and coordinates of point P.

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Answer: AP : PB = 11 : 18; P

(\frac{23}{29}, \frac{46}{29})

Let AP : PB $= k : 1$
P $= (\frac{7 k - 3}{k + 1}, \frac{- 4 k + 5}{k + 1})$
P lies on $y = 2 x$ : $- 4 k + 5 = 2 (7 k - 3)$ , so $18 k = 11$ , $k = \frac{11}{18}$
AP : PB $= 11 : 18$
P $= (\frac{11 \times 7 + 18 \times ( - 3 )}{29}, \frac{11 \times ( - 4 ) + 18 \times 5}{29}) = (\frac{23}{29}, \frac{46}{29})$

Q283 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/3/3

Determine the ratio in which the line $3 x + y - 9 = 0$ divides the line segment joining the points (1, 3) and (2, 5). Find the point of intersection.

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Answer: 3 : 2; point

(\frac{8}{5}, \frac{21}{5})

Let the line divide the segment in the ratio $k : 1$
Point $= (\frac{2 k + 1}{k + 1}, \frac{5 k + 3}{k + 1})$
It lies on the line: $3 (2 k + 1) + (5 k + 3) - 9 (k + 1) = 0$ , so $2 k - 3 = 0$ , $k = \frac{3}{2}$
Ratio $= 3 : 2$
Point $= (\frac{3 \times 2 + 2 \times 1}{5}, \frac{3 \times 5 + 2 \times 3}{5}) = (\frac{8}{5}, \frac{21}{5})$

Q293 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/5/1

A circle centered at $(2, 1)$ passes through the points $A (5, 6)$ and $B (- 3, K)$ . Find the value(s) of K. Hence find length of chord AB.

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Answer: K = 4 or K = −2; AB =

217

units (K = 4) or

82

units (K = −2)

Let centre $O (2, 1)$ . $O A^{2} = (5 - 2)^{2} + (6 - 1)^{2} = 9 + 25 = 34$
$O B^{2} = O A^{2}$ : $(- 3 - 2)^{2} + (K - 1)^{2} = 34$
$25 + (K - 1)^{2} = 34$ , so $(K - 1)^{2} = 9$ and $K - 1 = \pm 3$
$K = 4$ or $K = - 2$
For $K = 4$ : $A B = (5 + 3)^{2} + (6 - 4)^{2} = 68 = 217$
For $K = - 2$ : $A B = (5 + 3)^{2} + (6 + 2)^{2} = 128 = 82$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q29 (OR) (OR)3 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/5/1

Prove that the point P dividing the line segment joining the points $A (- 1, 7)$ and $B (4, - 3)$ in the ratio 3 : 2, lies on the line $x - 3 y = - 1$ . Also find length of PA and PB.

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Answer: P(2, 1) lies on the line; PA =

35

units, PB =

25

units

$P = (\frac{3 ( 4 ) + 2 ( - 1 )}{3 + 2}, \frac{3 ( - 3 ) + 2 ( 7 )}{3 + 2}) = (2, 1)$
Check: $x - 3 y = 2 - 3 = - 1$ . So P lies on the line $x - 3 y = - 1$ .
$P A = (2 + 1)^{2} + (1 - 7)^{2} = 9 + 36 = 35$
$P B = (4 - 2)^{2} + (- 3 - 1)^{2} = 4 + 16 = 25$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q293 marksShort AnswerCoordinate Geometry CBSE 2025 · Basic 430/4/1

If points A $(- 5, y)$ , B $(2, - 2)$ , C $(8, 4)$ and D $(x, 5)$ taken in order, form a parallelogram ABCD, then find the values of $x$ and $y$ . Hence, find lengths of sides of the parallelogram.

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Answer:

x = 1

y = - 1

; AB = CD =

52

units, BC = AD =

62

units

Diagonals of a parallelogram bisect each other, so the mid-points of AC and BD coincide.
Mid-point of AC $= (\frac{- 5 + 8}{2}, \frac{y + 4}{2})$ ; mid-point of BD $= (\frac{2 + x}{2}, \frac{- 2 + 5}{2})$ .
$\frac{2 + x}{2} = \frac{3}{2}$ gives $x = 1$ ; $\frac{y + 4}{2} = \frac{3}{2}$ gives $y = - 1$ .
AB $= (2 + 5)^{2} + (- 2 + 1)^{2} = 50 = 52$ units = CD.
BC $= (8 - 2)^{2} + (4 + 2)^{2} = 72 = 62$ units = AD.

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q29 (OR) (OR)3 marksShort AnswerCoordinate Geometry CBSE 2025 · Basic 430/4/1

A $(6, - 3)$ , B $(0, 5)$ and C $(- 2, 1)$ are vertices of $△$ ABC. Points P $(3, 1)$ and Q $(2, - 1)$ lie on sides AB and AC respectively. Check whether $\frac{A P}{P B} = \frac{A Q}{QC}$ .

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Answer: Yes,

\frac{A P}{P B} = \frac{A Q}{QC} = 1

AP $= (6 - 3)^{2} + (- 3 - 1)^{2} = 25 = 5$ ; PB $= (3 - 0)^{2} + (1 - 5)^{2} = 5$ .
So $\frac{A P}{P B} = 1$ .
AQ $= (6 - 2)^{2} + (- 3 + 1)^{2} = 20$ ; QC $= (2 + 2)^{2} + (- 1 - 1)^{2} = 20$ .
So $\frac{A Q}{QC} = 1$ .
Hence $\frac{A P}{P B} = \frac{A Q}{QC}$ .

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q283 marksShort AnswerCoordinate Geometry CBSE 2025 · Standard 30/1/1

Find the ratio in which the y-axis divides the line segment joining the points $(5, - 6)$ and $(- 1, - 4)$ . Also find the point of intersection.

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Answer: Ratio

5 : 1

; point of intersection

(0, - \frac{13}{3})

Let the y-axis divide the segment in the ratio $k : 1$ .
x-coordinate of the point $= \frac{k ( - 1 ) + 1 ( 5 )}{k + 1} = 0$ , so $k = 5$ . Ratio $= 5 : 1$ .
y-coordinate $= \frac{5 ( - 4 ) + 1 ( - 6 )}{5 + 1} = \frac{- 26}{6} = - \frac{13}{3}$ .
Point of intersection $= (0, - \frac{13}{3})$ .

Also asked in: 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q283 marksShort AnswerCoordinate Geometry CBSE 2025 · Standard 30/2/1

If the mid-point of the line segment joining the points $A (3, 4)$ and $B (k, 6)$ is $P (x, y)$ and $x + y - 10 = 0$ , then find the value of k.

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Answer:

k = 7

$x = \frac{3 + k}{2}$ , $y = \frac{4 + 6}{2} = 5$
$\frac{3 + k}{2} + 5 - 10 = 0$
$3 + k = 10$ , so $k = 7$

Also asked in: 2025 Standard 30/2/3

Q273 marksShort AnswerCoordinate Geometry CBSE 2025 · Standard 30/2/2

If $(a, b)$ is the mid-point of the line segment joining the points $A (10, - 6)$ and $B (k, 4)$ and $a - 2 b = 18$ , then find the value of k.

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Answer:

k = 22

$a = \frac{10 + k}{2}$ , $b = \frac{- 6 + 4}{2} = - 1$
$a - 2 b = 18$ gives $a + 2 = 18$ , so $a = 16$
$\frac{10 + k}{2} = 16$ , so $k = 22$

Q263 marksShort AnswerCoordinate Geometry CBSE 2025 · Standard 30/3/1

If the mid-point of the line segment joining the points $A (3, 4)$ and $B (k, 6)$ is $P (x, y)$ and $x + y - 10 = 0$ , find the value of k.

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Answer:

k = 7

$x = \frac{3 + k}{2}$ , $y = \frac{4 + 6}{2} = 5$ .
$x + y - 10 = 0$ gives $\frac{3 + k}{2} + 5 - 10 = 0$ .
$\frac{3 + k}{2} = 5$ , so $k = 7$ .

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q26 (OR) (OR)3 marksShort AnswerCoordinate Geometry CBSE 2025 · Standard 30/3/1

Find the coordinates of the points which divide the line segment joining $A (- 2, 2)$ and $B (2, 8)$ into four equal parts.

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Answer:

(- 1, \frac{7}{2})

(0, 5)

(1, \frac{13}{2})

Let P, Q, R divide AB in the ratios $1 : 3$ , $1 : 1$ and $3 : 1$ .
$P = (\frac{1 ( 2 ) + 3 ( - 2 )}{4}, \frac{1 ( 8 ) + 3 ( 2 )}{4}) = (- 1, \frac{7}{2})$ .
Q is the mid-point of AB: $Q = (\frac{- 2 + 2}{2}, \frac{2 + 8}{2}) = (0, 5)$ .
$R = (\frac{3 ( 2 ) + 1 ( - 2 )}{4}, \frac{3 ( 8 ) + 1 ( 2 )}{4}) = (1, \frac{13}{2})$ .

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q283 marksShort AnswerCoordinate Geometry CBSE 2025 · Standard 30/4/1

$P (x, y)$ , $Q (- 2, - 3)$ and $R (2, 3)$ are the vertices of a right triangle PQR right angled at P. Find the relationship between $x$ and $y$ . Hence, find all possible values of $x$ for which $y = 2$ .

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Answer:

x^{2} + y^{2} = 13

;

x = 3

x = - 3

Right angle at P: $P Q^{2} + P R^{2} = Q R^{2}$
$P Q^{2} = (x + 2)^{2} + (y + 3)^{2}$ , $P R^{2} = (x - 2)^{2} + (y - 3)^{2}$ , $Q R^{2} = 4^{2} + 6^{2} = 52$
$2 x^{2} + 2 y^{2} + 26 = 52 \Rightarrow x^{2} + y^{2} = 13$
Put $y = 2$ : $x^{2} = 9$ , so $x = \pm 3$

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q293 marksShort AnswerCoordinate Geometry CBSE 2025 · Standard 30/5/1

Find a relation between $x$ and $y$ such that $P (x, y)$ is equidistant from the points $A (3, 5)$ and $B (7, 1)$ . Hence, write the coordinates of the points on $x$ -axis and y-axis which are equidistant from points A and B.

Show answer & solution

Answer:

x - y = 2

; on x-axis (2, 0), on y-axis (0,

- 2

)

$P A^{2} = P B^{2}$ : $(x - 3)^{2} + (y - 5)^{2} = (x - 7)^{2} + (y - 1)^{2}$
$- 6 x - 10 y + 34 = - 14 x - 2 y + 50$
$8 x - 8 y = 16 \Rightarrow x - y = 2$
On x-axis ( $y = 0$ ): $x = 2$ , point (2, 0)
On y-axis ( $x = 0$ ): $y = - 2$ , point (0, $- 2$ )

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q293 marksShort AnswerCoordinate Geometry CBSE 2025 · Standard 30/6/1

If the points $A (6, 1)$ , $B (p, 2)$ , $C (9, 4)$ and $D (7, q)$ are the vertices of a parallelogram ABCD, then find the values of $p$ and $q$ . Hence, check whether ABCD is a rectangle or not.

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Answer:

p = 8

q = 3

; ABCD is not a rectangle

Diagonals of a parallelogram bisect each other: midpoint of AC = midpoint of BD.
Midpoint of AC $= (\frac{15}{2}, \frac{5}{2})$ ; midpoint of BD $= (\frac{p + 7}{2}, \frac{2 + q}{2})$
$p + 7 = 15 \Rightarrow p = 8$ ; $2 + q = 5 \Rightarrow q = 3$
$A C = 3^{2} + 3^{2} = 32$ ; $B D = (7 - 8)^{2} + (3 - 2)^{2} = 2$
Diagonals are not equal, so ABCD is not a rectangle.

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q303 marksShort AnswerCoordinate Geometry CBSE 2024 · Basic 430/1/2

The line AB intersects x-axis at A and y-axis at B. The point $P (2, - 3)$ lies on AB such that AP : PB = 3 : 1. Find the co-ordinates of A and B.

Diagram for CBSE 2024 Class 10 Maths question 30

Show answer & solution

Answer: A(8, 0), B(0,

- 4

)

Let A be $(a, 0)$ and B be $(0, b)$ .
P divides AB in the ratio 3 : 1, so $P = (\frac{3 \times 0 + 1 \times a}{4}, \frac{3 \times b + 1 \times 0}{4})$ .
$\frac{a}{4} = 2$ gives $a = 8$ ; $\frac{3 b}{4} = - 3$ gives $b = - 4$ .
So A(8, 0) and B(0, $- 4$ ).

Q283 marksShort AnswerCoordinate Geometry CBSE 2024 · Basic 430/1/3

Find the co-ordinates of the points of trisection of the line segment joining the points $(- 2, 2)$ and $(7, - 4)$ .

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Answer: (1, 0) and (4,

- 2

)

The points of trisection divide the segment in the ratios 1 : 2 and 2 : 1.
1 : 2: $(\frac{1 \times 7 + 2 \times ( - 2 )}{3}, \frac{1 \times ( - 4 ) + 2 \times 2}{3}) = (1, 0)$ .
2 : 1: $(\frac{2 \times 7 + 1 \times ( - 2 )}{3}, \frac{2 \times ( - 4 ) + 1 \times 2}{3}) = (4, - 2)$ .

Q313 marksShort AnswerCoordinate Geometry CBSE 2024 · Basic 430/5/1

Find the coordinates of the points of trisection of the line segment joining the points $A (5, - 3)$ and $B (- 4, 3)$ .

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Answer:

(2, - 1)

and

(- 1, 1)

Let P and Q divide AB in the ratios $1 : 2$ and $2 : 1$ .
$P = (\frac{1 \times ( - 4 ) + 2 \times 5}{3}, \frac{1 \times 3 + 2 \times ( - 3 )}{3}) = (2, - 1)$ .
$Q = (\frac{2 \times ( - 4 ) + 1 \times 5}{3}, \frac{2 \times 3 + 1 \times ( - 3 )}{3}) = (- 1, 1)$ .

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q263 marksShort AnswerCoordinate Geometry CBSE 2024 · Standard 30/1/1

Find the ratio in which the point $(\frac{8}{5}, y)$ divides the line segment joining the points $(1, 2)$ and $(2, 3)$ . Also, find the value of $y$ .

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Answer: Ratio

3 : 2

;

y = \frac{13}{5}

Let the point divide the segment in the ratio $k : 1$ .
$x$ -coordinate: $\frac{2 k + 1}{k + 1} = \frac{8}{5}$ , so $10 k + 5 = 8 k + 8$ and $k = \frac{3}{2}$ .
Ratio $= 3 : 2$ .
$y = \frac{3 \times 3 + 2 \times 2}{3 + 2} = \frac{13}{5}$ .

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q26 (OR) (OR)3 marksShort AnswerCoordinate Geometry CBSE 2024 · Standard 30/1/1

ABCD is a rectangle formed by the points A $(- 1, - 1)$ , B $(- 1, 6)$ , C $(3, 6)$ and D $(3, - 1)$ . P, Q, R and S are mid–points of sides AB, BC, CD and DA respectively. Show that diagonals of the quadrilateral PQRS bisect each other.

Show answer & solution

Answer: Mid-points of PR and QS are both

(1, \frac{5}{2})

, so the diagonals bisect each other.

P (mid-point of AB) $= (- 1, \frac{5}{2})$ , Q (mid-point of BC) $= (1, 6)$ .
R (mid-point of CD) $= (3, \frac{5}{2})$ , S (mid-point of DA) $= (1, - 1)$ .
Mid-point of PR $= (\frac{- 1 + 3}{2}, \frac{5}{2}) = (1, \frac{5}{2})$ .
Mid-point of QS $= (\frac{1 + 1}{2}, \frac{6 - 1}{2}) = (1, \frac{5}{2})$ .
The diagonals PR and QS have the same mid-point, so they bisect each other.

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q293 marksShort AnswerCoordinate Geometry CBSE 2024 · Standard 30/3/1

Find the ratio in which the line segment joining the points (5, 3) and ( $- 1$ , 6) is divided by Y-axis.

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Answer: 5 : 1 (the point of division is

(0, \frac{11}{2})

)

Let the y-axis divide the segment in the ratio $k : 1$ at the point $(0, y)$ .
$x$ -coordinate: $\frac{k ( - 1 ) + 1 ( 5 )}{k + 1} = 0 \Rightarrow k = 5$ .
The ratio is 5 : 1.
$y = \frac{5 ( 6 ) + 1 ( 3 )}{6} = \frac{33}{6} = \frac{11}{2}$ , so the point is $(0, \frac{11}{2})$ .

Also asked in: 2024 Standard 30/3/3

Q29 (OR) (OR)3 marksShort AnswerCoordinate Geometry CBSE 2024 · Standard 30/3/1

P( $- 2$ , 5) and Q(3, 2) are two points. Find the coordinates of the point R on line segment PQ such that PR = 2QR.

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Answer: R

(\frac{4}{3}, 3)

PR = 2QR, so R divides PQ internally in the ratio 2 : 1.
$x = \frac{2 ( 3 ) + 1 ( - 2 )}{3} = \frac{4}{3}$ .
$y = \frac{2 ( 2 ) + 1 ( 5 )}{3} = 3$ .
R is $(\frac{4}{3}, 3)$ .

Also asked in: 2024 Standard 30/3/3

Q283 marksShort AnswerCoordinate Geometry CBSE 2024 · Standard 30/3/2

In what ratio does the X-axis divides the line segment joining the points(2, $- 3$ ) and (5, 6) ? Also, find the coordinates of the point of intersection.

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Answer: 1 : 2; point of intersection (3, 0)

Let the x-axis divide the segment in the ratio $k : 1$ at $(x, 0)$ .
$y$ -coordinate: $\frac{6 k - 3}{k + 1} = 0 \Rightarrow k = \frac{1}{2}$ , so the ratio is 1 : 2.
$x = \frac{1 ( 5 ) + 2 ( 2 )}{1 + 2} = \frac{9}{3} = 3$ .
Point of intersection: (3, 0).

Q28 (OR) (OR)3 marksShort AnswerCoordinate Geometry CBSE 2024 · Standard 30/3/2

Find the length of the median AD of $△ A B C$ having vertices A(0, $- 1$ ), B(2, 1) and C(0, 3).

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Answer: AD =

10

units

D is the mid-point of BC: D = $(\frac{2 + 0}{2}, \frac{1 + 3}{2}) = (1, 2)$ .
AD = $(1 - 0)^{2} + (2 + 1)^{2} = 1 + 9 = 10$ units.

Q273 marksShort AnswerCoordinate Geometry CBSE 2023 · Basic 430/1/1

Show that the points $A (1, 7)$ , $B (4, 2)$ $C (- 1, - 1)$ and $D (- 4, 4)$ are vertices of the square ABCD.

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Answer: Proved.

$A B = (4 - 1)^{2} + (2 - 7)^{2} = 9 + 25 = 34$
$B C = (- 1 - 4)^{2} + (- 1 - 2)^{2} = 25 + 9 = 34$
$C D = (- 4 + 1)^{2} + (4 + 1)^{2} = 9 + 25 = 34$
$D A = (1 + 4)^{2} + (7 - 4)^{2} = 25 + 9 = 34$
Diagonals: $A C = (- 1 - 1)^{2} + (- 1 - 7)^{2} = 68$ , $B D = (- 4 - 4)^{2} + (4 - 2)^{2} = 68$
All four sides are equal and the diagonals are equal, so ABCD is a square.

Also asked in: 2023 Basic 430/1/2

Q303 marksShort AnswerCoordinate Geometry CBSE 2023 · Basic 430/1/3

If the point $Q (0, 1)$ is equidistant from the points $P (5, - 3)$ and $R (x, 6)$ ; find the value of $x$ . Also, find the distance PR.

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Answer:

x = \pm 4

; PR

= 82

units (when

x = 4

) or

92

units (when

x = - 4

)

$Q P^{2} = (5 - 0)^{2} + (- 3 - 1)^{2} = 25 + 16 = 41$
$Q R^{2} = (x - 0)^{2} + (6 - 1)^{2} = x^{2} + 25$
QP = QR gives $x^{2} + 25 = 41$ , so $x^{2} = 16$ and $x = \pm 4$ .
For $x = 4$ : $P R = (4 - 5)^{2} + (6 + 3)^{2} = 1 + 81 = 82$ units
For $x = - 4$ : $P R = (- 4 - 5)^{2} + (6 + 3)^{2} = 81 + 81 = 92$ units

Q313 marksShort AnswerCoordinate Geometry CBSE 2023 · Basic 430/2/1

Prove that the points $A (- 1, 0)$ , $B (3, 1)$ , $C (2, 2)$ and $D (- 2, 1)$ are the vertices of a parallelogram ABCD. Is it also a rectangle ?

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Answer: ABCD is a parallelogram; it is not a rectangle.

Mid-point of AC $= (\frac{- 1 + 2}{2}, \frac{0 + 2}{2}) = (\frac{1}{2}, 1)$
Mid-point of BD $= (\frac{3 - 2}{2}, \frac{1 + 1}{2}) = (\frac{1}{2}, 1)$
The diagonals bisect each other, so ABCD is a parallelogram.
$A C = 3^{2} + 2^{2} = 13$ , $B D = (- 5)^{2} + 0^{2} = 5$
The diagonals are not equal, so ABCD is not a rectangle.

Also asked in: 2023 Basic 430/2/2

Q263 marksShort AnswerCoordinate Geometry CBSE 2023 · Basic 430/2/3

Show that the points $A (- 3, 2)$ , $B (- 5, - 5)$ , $C (2, - 3)$ and $D (4, 4)$ are vertices of a rhombus ABCD. Is it also a square ?

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Answer: ABCD is a rhombus; it is not a square.

$A B = (- 2)^{2} + (- 7)^{2} = 53$
$B C = 7^{2} + 2^{2} = 53$
$C D = 2^{2} + 7^{2} = 53$
$D A = (- 7)^{2} + (- 2)^{2} = 53$
All sides are equal, so ABCD is a rhombus.
Diagonals: $A C = 5^{2} + (- 5)^{2} = 50$ , $B D = 9^{2} + 9^{2} = 162$
The diagonals are not equal, so ABCD is not a square.

Q293 marksShort AnswerCoordinate Geometry CBSE 2023 · Basic 430/5/1

Determine the ratio in which the point P(a, $-$ 2) divides the line segment joining the points A( $-$ 4, 3) and B(2, $-$ 4). Also, find the value of ‘a’.

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Answer: 5 : 2,

a = \frac{2}{7}

Let P divide AB in the ratio $k : 1$ .
y-coordinate: $\frac{- 4 k + 3}{k + 1} = - 2$ , so $- 4 k + 3 = - 2 k - 2$ , $k = \frac{5}{2}$ .
Ratio = 5 : 2.
x-coordinate: $a = \frac{5 \times 2 + 2 \times ( - 4 )}{5 + 2} = \frac{2}{7}$ .

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q29 (OR) (OR)3 marksShort AnswerCoordinate Geometry CBSE 2023 · Basic 430/5/1

In the given figure, in $△ A B C$ points D and E are mid-points of sides BC and AC respectively. If given vertices are A(4, $-$ 2), B(2, $-$ 2) and C( $-$ 6, $-$ 7), then verify the result DE = $\frac{1}{2}$ AB.

Diagram for CBSE 2023 Class 10 Maths question 29 (OR)

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Answer: Verified: DE = 1 unit, AB = 2 units.

D = midpoint of BC $= (\frac{2 - 6}{2}, \frac{- 2 - 7}{2}) = (- 2, - \frac{9}{2})$ .
E = midpoint of AC $= (\frac{4 - 6}{2}, \frac{- 2 - 7}{2}) = (- 1, - \frac{9}{2})$ .
DE $= (- 1 + 2)^{2} + 0^{2} = 1$ .
AB $= (4 - 2)^{2} + 0^{2} = 2$ .
So DE $= \frac{1}{2}$ AB. Verified.

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q293 marksShort AnswerCoordinate Geometry CBSE 2023 · Basic 430/6/1

Find the co-ordinates of the points of trisection of the line-segment joining the points (5, 3) and (4, 5).

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Answer:

(\frac{14}{3}, \frac{11}{3})

and

(\frac{13}{3}, \frac{13}{3})

Let A(5, 3), B(4, 5). The points of trisection divide AB in the ratios 1 : 2 and 2 : 1.
1 : 2: $(\frac{1 \times 4 + 2 \times 5}{3}, \frac{1 \times 5 + 2 \times 3}{3}) = (\frac{14}{3}, \frac{11}{3})$
2 : 1: $(\frac{2 \times 4 + 1 \times 5}{3}, \frac{2 \times 5 + 1 \times 3}{3}) = (\frac{13}{3}, \frac{13}{3})$

Also asked in: 2023 Basic 430/6/3

Q273 marksShort AnswerCoordinate Geometry CBSE 2023 · Basic 430/6/2

If A and B are (–2, –2) and (2, –4) respectively; then find the co-ordinates of the point P such that $\frac{A P}{A B} = \frac{3}{7}$ .

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Answer:

P (- \frac{2}{7}, - \frac{20}{7})

$\frac{A P}{A B} = \frac{3}{7}$ , so $A P : P B = 3 : 4$ (P lies on AB).
$x = \frac{3 \times 2 + 4 \times ( - 2 )}{7} = \frac{- 2}{7}$
$y = \frac{3 \times ( - 4 ) + 4 \times ( - 2 )}{7} = \frac{- 20}{7}$
$P = (- \frac{2}{7}, - \frac{20}{7})$ .

Q283 marksShort AnswerCoordinate Geometry CBSE 2023 · Standard 30/1/1

If (–5, 3) and (5, 3) are two vertices of an equilateral triangle, then find co-ordinates of the third vertex, given that origin lies inside the triangle. (Take $3 = 1.7$ )

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Answer: (0, –5.5)

Let A(–5, 3), B(5, 3); AB = 10.
The third vertex C is equidistant from A and B, so it lies on x = 0: C(0, y).
$A C^{2} = 25 + (y - 3)^{2} = 100$ , so $(y - 3)^{2} = 75$ , $y = 3 \pm 53$ .
Origin is inside the triangle, so C is below AB: $y = 3 - 53 = 3 - 8.5 = - 5.5$ .
Third vertex = (0, –5.5)