CBSE Maths › Class 10 PYQs › Coordinate Geometry

Coordinate Geometry: 2 marks Questions (CBSE Class 10)

57 different 2 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)

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Basic 430/4/1

A (– 2, 3) and B (4, 1) are end points of the diameter of a semi-circle. The semi-circle intersects y-axis at point P. Find the co-ordinates of the point P.

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Answer: P(0, 5) or P(0, – 1)

Centre = midpoint of AB = (1, 2); radius = $(4 - 1)^{2} + (1 - 2)^{2} = 10$ .
Let P = (0, y). Then $(0 - 1)^{2} + (y - 2)^{2} = 10$ .
$(y - 2)^{2} = 9 \Rightarrow y = 5$ or $y = - 1$ .
So P is (0, 5) or (0, – 1), depending on which side of AB the semi-circle lies.

Also asked in: 2026 Basic 430/4/2

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Basic 430/4/3

The vertices of a $△ A B C$ are A(–1, 3), B(2, –3) and C(4, 5). Find the coordinates of a point P on median AD such that AP : PD = 2 : 3.

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Answer: P

(\frac{3}{5}, \frac{11}{5})

D is the midpoint of BC: D = $(\frac{2 + 4}{2}, \frac{- 3 + 5}{2}) = (3, 1)$ .
P divides AD in the ratio 2 : 3.
P = $(\frac{2 ( 3 ) + 3 ( - 1 )}{5}, \frac{2 ( 1 ) + 3 ( 3 )}{5}) = (\frac{3}{5}, \frac{11}{5})$ .

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Basic 430/5/1

If A(a, 0), B(1, 1) and C(0, b) form a triangle, right angled at B when joined, then establish a relation between a and b.

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Answer:

a + b = 2

Right angle at B, so $A B^{2} + B C^{2} = A C^{2}$ .
$A B^{2} = (1 - a)^{2} + 1$ , $B C^{2} = 1 + (b - 1)^{2}$ , $A C^{2} = a^{2} + b^{2}$
$1 - 2 a + a^{2} + 1 + 1 + b^{2} - 2 b + 1 = a^{2} + b^{2}$
$4 - 2 a - 2 b = 0$ , so $a + b = 2$

Also asked in: 2026 Basic 430/5/2

Q252 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Basic 430/5/3

In the given figure, PQ is a tangent to a circle with centre O(−5, 3). If coordinates of P and Q are (3, 1) and (0, 6) respectively, then using distance formula, show that PQ $⊥$ OQ.

Diagram for CBSE 2026 Class 10 Maths question 25

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Answer: Proved:

O Q^{2} + P Q^{2} = 34 + 34 = 68 = O P^{2}

, so

∠ O QP = 9 0^{\circ}

$O Q^{2} = (0 + 5)^{2} + (6 - 3)^{2} = 25 + 9 = 34$
$P Q^{2} = (3 - 0)^{2} + (1 - 6)^{2} = 9 + 25 = 34$
$O P^{2} = (3 + 5)^{2} + (1 - 3)^{2} = 64 + 4 = 68$
$O Q^{2} + P Q^{2} = O P^{2}$ , so by the converse of Pythagoras theorem $∠ O QP = 9 0^{\circ}$ , i.e. PQ $⊥$ OQ.

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/1/1

The coordinates of the centre of a circle are $(x - 7, 2 x)$ . Find the value(s) of ' $x$ ', if the circle passes through the point $(- 9, 11)$ and has radius $52$ units.

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Answer:

x = 3

x = 5

Distance of $(- 9, 11)$ from the centre equals the radius.
$(x - 7 + 9)^{2} + (2 x - 11)^{2} = (52)^{2} = 50$
$(x + 2)^{2} + (2 x - 11)^{2} = 50$
$x^{2} + 4 x + 4 + 4 x^{2} - 44 x + 121 = 50$
$5 x^{2} - 40 x + 75 = 0$ , i.e. $x^{2} - 8 x + 15 = 0$
$(x - 3) (x - 5) = 0$ , so $x = 3$ or $x = 5$ .

Also asked in: 2026 Standard 30/1/2

Q212 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/1/3

Do the points P $(1, 0)$ , Q $(- 5, 0)$ and R $(- 2, 5)$ form a triangle ? If so, name the type of triangle formed.

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Answer: Yes; an isosceles triangle (

P R = QR = 34

P Q = 6

$P Q = (1 + 5)^{2} + 0^{2} = 6$ .
$QR = (- 2 + 5)^{2} + (5 - 0)^{2} = 9 + 25 = 34$ .
$P R = (- 2 - 1)^{2} + (5 - 0)^{2} = 9 + 25 = 34$ .
The sum of any two sides exceeds the third (e.g. $234 > 6$ ; also P, Q lie on the $x$ -axis but R does not), so the points are not collinear and form a triangle.
Since $QR = P R$ , it is an isosceles triangle.

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/2/1

If the points A(4, 5), B(m, 6), C(4, 3) and D(1, n) taken in this order are the vertices of a parallelogram ABCD, then find the values of m and n.

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Answer: m = 7, n = 2

The diagonals of a parallelogram bisect each other, so mid-point of AC = mid-point of BD.
Mid-point of AC $= (\frac{4 + 4}{2}, \frac{5 + 3}{2}) = (4, 4)$
Mid-point of BD $= (\frac{m + 1}{2}, \frac{6 + n}{2})$
$\frac{m + 1}{2} = 4 \Rightarrow m = 7$ ; $\frac{6 + n}{2} = 4 \Rightarrow n = 2$

Also asked in: 2026 Standard 30/2/2

Q212 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/2/3

If the distance between the points (4, p) and (1, 0) is 5, what is the value of p ?

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Answer:

p = \pm 4

By the distance formula, $(4 - 1)^{2} + (p - 0)^{2} = 5^{2}$ .
$9 + p^{2} = 25$
$p^{2} = 16$ , so $p = 4$ or $p = - 4$ .

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/3/1

Diagonals AC and BD of square ABCD intersect at P. Coordinates of points B and D are (9, – 2) and (1, 6) respectively.
(i) Find the co-ordinates of point P.
(ii) Find the length of the side of the square.

Diagram for CBSE 2026 Class 10 Maths question 23

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Answer: (i) P(5, 2) (ii) 8 units

(i) Diagonals of a square bisect each other, so P is the midpoint of BD
P $= (\frac{9 + 1}{2}, \frac{- 2 + 6}{2}) = (5, 2)$
(ii) BD $= (9 - 1)^{2} + (- 2 - 6)^{2} = 128 = 82$
Diagonal $= 2 \times$ side, so side $= \frac{8 2}{2} = 8$ units

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

Q23 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/3/1

Find the coordinates of a point on the line $x + y = 5$ which is equidistant from (6, 4) and (5, 2).

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Answer:

(- \frac{3}{2}, \frac{13}{2})

Let the point be $(x, y)$
$(x - 6)^{2} + (y - 4)^{2} = (x - 5)^{2} + (y - 2)^{2}$
$- 12 x + 36 - 8 y + 16 = - 10 x + 25 - 4 y + 4$ , so $2 x + 4 y = 23$
With $y = 5 - x$ : $2 x + 20 - 4 x = 23$ , so $x = - \frac{3}{2}$ , $y = \frac{13}{2}$
Point: $(- \frac{3}{2}, \frac{13}{2})$

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

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/5/1

Vertices of a right triangle ABC with $∠ B = 9 0^{\circ}$ are $A (3, 4)$ , $B (1, 1)$ and $C (- 8, 7)$ . Find the value of tan A.

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Answer:

tan A = 3

$A B = (3 - 1)^{2} + (4 - 1)^{2} = 4 + 9 = 13$
$B C = (1 + 8)^{2} + (1 - 7)^{2} = 81 + 36 = 117 = 313$
In right $△ A B C$ with $∠ B = 9 0^{\circ}$ : $tan A = \frac{B C}{A B} = \frac{3 13}{13} = 3$

Also asked in: 2026 Standard 30/5/2

Q22 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/5/1

Using distance formula, prove that the points $A (2, 3)$ , $B (- 7, 0)$ and $C (- 1, 2)$ are collinear.

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Answer: Proved.

$A B = (2 + 7)^{2} + (3 - 0)^{2} = 81 + 9 = 90 = 310$
$B C = (- 1 + 7)^{2} + (2 - 0)^{2} = 36 + 4 = 40 = 210$
$A C = (2 + 1)^{2} + (3 - 2)^{2} = 9 + 1 = 10$
$B C + C A = 210 + 10 = 310 = A B$
Hence A, B, C are collinear (C lies between A and B).

Also asked in: 2026 Standard 30/5/2

Q212 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/5/3

In the given figure, point D divides the side BC of $△ A B C$ in the ratio 1 : 2. Find length AD.

Diagram for CBSE 2026 Class 10 Maths question 21

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Answer: AD =

\frac{130}{3}

units (≈ 3.80 units)

From the figure: $A (1, 5)$ , $B (- 2, 1)$ , $C (4, 2)$ ; BD : DC = 1 : 2.
$D = (\frac{1 ( 4 ) + 2 ( - 2 )}{1 + 2}, \frac{1 ( 2 ) + 2 ( 1 )}{1 + 2}) = (0, \frac{4}{3})$
$A D = (1 - 0)^{2} + (5 - \frac{4}{3})^{2} = 1 + \frac{121}{9} = \frac{130}{9} = \frac{130}{3}$ units

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Basic 430/4/1

Point P $(x, 0)$ divides the line segment joining the points $(2, 8)$ and $(- 3, - 5)$ in a certain ratio. Find the ratio and hence find the value of $x$ .

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Answer: 8 : 5;

x = - \frac{14}{13}

Let P divide the segment in the ratio $k : 1$ .
$y$ -coordinate: $\frac{- 5 k + 8}{k + 1} = 0$ , so $k = \frac{8}{5}$ ; the ratio is 8 : 5.
$x = \frac{8 \times ( - 3 ) + 5 \times 2}{8 + 5} = \frac{- 24 + 10}{13} = - \frac{14}{13}$ .

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Basic 430/4/2

Find the ratio in which point P $(- 1, m)$ divides the line segment joining the points A $(2, 5)$ and B $(- 5, - 2)$ . Hence, find the value of $m$ .

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Answer: 3 : 4;

m = 2

Let P divide AB in the ratio $k : 1$ .
$x$ -coordinate: $\frac{- 5 k + 2}{k + 1} = - 1$ , so $- 5 k + 2 = - k - 1$ and $k = \frac{3}{4}$ ; ratio 3 : 4.
$m = \frac{3 \times ( - 2 ) + 4 \times 5}{3 + 4} = \frac{14}{7} = 2$ .

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Basic 430/4/3

Find the ratio in which the segment joining the points $(2, - 5)$ and $(5, 3)$ is divided by $x$ -axis. Also, find coordinates of the point on $x$ -axis.

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Answer: 5 : 3;

(\frac{31}{8}, 0)

Let the $x$ -axis divide the segment in the ratio $k : 1$ at a point $(x, 0)$ .
$\frac{3 k - 5}{k + 1} = 0$ , so $k = \frac{5}{3}$ ; ratio 5 : 3.
$x = \frac{5 \times 5 + 3 \times 2}{5 + 3} = \frac{31}{8}$ , so the point is $(\frac{31}{8}, 0)$ .

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Basic 430/5/1

Find the coordinates of the points of trisection of line segment joining the points (–4, 1) and (6, 5).

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Answer:

(- \frac{2}{3}, \frac{7}{3})

and

(\frac{8}{3}, \frac{11}{3})

Let A(–4, 1), B(6, 5). The points of trisection divide AB in the ratios 1 : 2 and 2 : 1.
For 1 : 2: $(\frac{1 \times 6 + 2 \times ( - 4 )}{3}, \frac{1 \times 5 + 2 \times 1}{3}) = (- \frac{2}{3}, \frac{7}{3})$ .
For 2 : 1: $(\frac{2 \times 6 + 1 \times ( - 4 )}{3}, \frac{2 \times 5 + 1 \times 1}{3}) = (\frac{8}{3}, \frac{11}{3})$ .

Also asked in: 2025 Basic 430/5/3

Q252 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Basic 430/5/1

Using distance formula, prove that the points (1, 5), (2, 3) and (3, 1) are collinear.

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Answer: Proved.

Let A(1, 5), B(2, 3), C(3, 1).
$A B = 1^{2} + (- 2)^{2} = 5$ , $B C = 1^{2} + (- 2)^{2} = 5$ .
$A C = 2^{2} + (- 4)^{2} = 20 = 25$ .
$A B + B C = 25 = A C$ , so A, B, C are collinear.

Also asked in: 2025 Basic 430/5/2

Q212 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Basic 430/5/2

Establish a relation between $x$ and y such that point ( $x$ , y) is equidistant from points (–2, 5) and (3, 9).

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Answer:

10 x + 8 y = 61

(i.e.

10 x + 8 y - 61 = 0

)

Let P( $x$ , y) be equidistant from A(–2, 5) and B(3, 9), so $P A^{2} = P B^{2}$ .
$(x + 2)^{2} + (y - 5)^{2} = (x - 3)^{2} + (y - 9)^{2}$ .
$x^{2} + 4 x + 4 + y^{2} - 10 y + 25 = x^{2} - 6 x + 9 + y^{2} - 18 y + 81$ .
$4 x - 10 y + 29 = - 6 x - 18 y + 90$ , so $10 x + 8 y - 61 = 0$ .

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Basic 430/5/3

Using distance formula, show that the points (–1, 3), (6, 2) and (3, –1) are vertices of a right-angled triangle.

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Answer: Proved (right angle at (3, –1)).

Let A(–1, 3), B(6, 2), C(3, –1).
$A B^{2} = (6 + 1)^{2} + (2 - 3)^{2} = 49 + 1 = 50$ .
$B C^{2} = (3 - 6)^{2} + (- 1 - 2)^{2} = 9 + 9 = 18$ .
$A C^{2} = (3 + 1)^{2} + (- 1 - 3)^{2} = 16 + 16 = 32$ .
$B C^{2} + A C^{2} = 18 + 32 = 50 = A B^{2}$ , so by the converse of Pythagoras theorem, $△ A B C$ is right-angled at C.

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Standard 30/1/1

The coordinates of the centre of a circle are $(2 a, a - 7)$ . Find the value(s) of 'a' if the circle passes through the point $(11, - 9)$ and has diameter $102$ units.

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Answer:

a = 3

a = 5

Radius $= 52$ , so (radius) $^{2} = 50$ .
Distance from centre $(2 a, a - 7)$ to $(11, - 9)$ equals the radius:
$(11 - 2 a)^{2} + (- 9 - a + 7)^{2} = 50$
$(11 - 2 a)^{2} + (a + 2)^{2} = 50$
$121 - 44 a + 4 a^{2} + a^{2} + 4 a + 4 = 50$
$5 a^{2} - 40 a + 75 = 0$ , i.e. $a^{2} - 8 a + 15 = 0$
$(a - 3) (a - 5) = 0$ , so $a = 3$ or $a = 5$ .

Also asked in: 2025 Standard 30/1/2

Q252 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Standard 30/1/3

Find the length of the median through the vertex $B$ of $△ A B C$ with vertices $A (9, - 2)$ , $B (- 3, 7)$ and $C (- 1, 10)$ .

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Answer:

58

units

The median through $B$ joins $B$ to the mid-point $D$ of $A C$ .
$D = (\frac{9 + ( - 1 )}{2}, \frac{- 2 + 10}{2}) = (4, 4)$ .
$B D = (4 + 3)^{2} + (4 - 7)^{2} = 49 + 9 = 58$ units.

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Standard 30/4/1

Find the coordinates of the point C which lies on the line AB produced such that $A C = 2 B C$ , where coordinates of points A and B are $(- 1, 7)$ and $(4, - 3)$ respectively.

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Answer:

C (9, - 13)

C lies on AB produced beyond B and $A C = 2 B C$ , so $A B = A C - B C = B C$ : B is the mid-point of AC.
Let $C = (x, y)$ . Then $\frac{- 1 + x}{2} = 4$ and $\frac{7 + y}{2} = - 3$
$x = 9$ , $y = - 13$ , so $C = (9, - 13)$

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

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Standard 30/5/1

The coordinates of the end points of the line segment AB are $A (- 2, - 2)$ and $B (2, - 4)$ . P is the point on AB such that $B P = \frac{4}{7} A B$ . Find the coordinates of point P.

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Answer:

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

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

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

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2025 · Standard 30/6/1

Prove that abscissa of a point P which is equidistant from points with coordinates $A (7, 1)$ and $B (3, 5)$ is 2 more than its ordinate.

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Answer: Proved.

Let $P (x, y)$ . $P A = P B \Rightarrow P A^{2} = P B^{2}$
$(x - 7)^{2} + (y - 1)^{2} = (x - 3)^{2} + (y - 5)^{2}$
$- 14 x + 49 - 2 y + 1 = - 6 x + 9 - 10 y + 25$
$- 8 x + 8 y + 16 = 0 \Rightarrow x - y = 2$
So $x = y + 2$ : the abscissa is 2 more than the ordinate.

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

Q212 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Basic 430/3/1

If $Q (0, 2)$ is equidistant from $P (5, - 3)$ and $R (x, 7)$ , find the value(s) of $x$ .

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Answer:

x = \pm 5

$QP = QR$ , so $Q P^{2} = Q R^{2}$ .
$Q P^{2} = (5 - 0)^{2} + (- 3 - 2)^{2} = 25 + 25 = 50$ .
$Q R^{2} = (x - 0)^{2} + (7 - 2)^{2} = x^{2} + 25$ .
$x^{2} + 25 = 50$ , so $x^{2} = 25$ and $x = \pm 5$ .

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

Q21 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Basic 430/3/1

If $A (1, 1)$ and $B (7, 9)$ are the end points of a diameter of a circle, then find the co-ordinates of the centre of the circle.

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Answer: (4, 5)

The centre is the mid-point of the diameter AB.
Centre $= (\frac{1 + 7}{2}, \frac{1 + 9}{2}) = (4, 5)$ .

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

Q242 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Basic 430/3/1

Find the ratio in which the Y-axis divides the line segment joining the points $A (5, - 6)$ and $B (- 1, - 4)$ . 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 AB in the ratio $k : 1$ . The point is $(\frac{- k + 5}{k + 1}, \frac{- 4 k - 6}{k + 1})$ .
On the Y-axis the x-coordinate is 0, so $- k + 5 = 0$ and $k = 5$ . Ratio $= 5 : 1$ .
y-coordinate $= \frac{- 20 - 6}{6} = - \frac{13}{3}$ .
Point of intersection $= (0, - \frac{13}{3})$ .

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Basic 430/3/2

Find a point which is equidistant from the points $A (- 1, 5)$ and $B (2, 1)$ . How many such points are there ?

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Answer: For example

(\frac{1}{2}, 3)

(the mid-point of AB); there are infinitely many such points, all on the line

6 x - 8 y + 21 = 0

The mid-point of AB is equidistant from A and B: $(\frac{- 1 + 2}{2}, \frac{5 + 1}{2}) = (\frac{1}{2}, 3)$ .
In general $P (x, y)$ is equidistant if $(x + 1)^{2} + (y - 5)^{2} = (x - 2)^{2} + (y - 1)^{2}$ .
This simplifies to $6 x - 8 y + 21 = 0$ , the perpendicular bisector of AB.
Every point on this line works, so there are infinitely many such points.

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Basic 430/3/3

Point $P (x, y)$ divides the line segment joining the points $A (- 1, 3)$ and $B (9, 8)$ such that $A P : P B = k : 1$ . If the co-ordinates of P are such that $x = y$ , then find the value of k.

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Answer:

k = 4

By the section formula, $x = \frac{9 k - 1}{k + 1}$ and $y = \frac{8 k + 3}{k + 1}$ .
$x = y$ gives $9 k - 1 = 8 k + 3$ , so $k = 4$ .
Check: P = (7, 7).

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Basic 430/5/1

Find the ratio in which the point $(3, y)$ , divides the line segment joining the points $(- 2, - 5)$ and $(6, 3)$ . Also, find the value of y.

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Answer: Ratio

5 : 3

;

y = 0

Let the point divide the segment in the ratio $k : 1$ .
x-coordinate: $\frac{6 k - 2}{k + 1} = 3$ , so $6 k - 2 = 3 k + 3$ , giving $k = \frac{5}{3}$ .
So the ratio is $5 : 3$ .
$y = \frac{5 \times 3 + 3 \times ( - 5 )}{5 + 3} = \frac{15 - 15}{8} = 0$ .

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

Q22 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Basic 430/5/1

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

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Answer:

(0, 9)

Let the point be $P (0, y)$ . Then $P A^{2} = P B^{2}$ .
$36 + (y - 5)^{2} = 16 + (y - 3)^{2}$ .
$36 + y^{2} - 10 y + 25 = 16 + y^{2} - 6 y + 9$ .
$61 - 10 y = 25 - 6 y$ , so $4 y = 36$ and $y = 9$ .
The point is $(0, 9)$ .

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

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Standard 30/2/1

Find the type of triangle ABC formed whose vertices are A(1, 0), B( $- 5$ , 0) and C( $- 2$ , 5).

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Answer: Isosceles triangle (AC = BC =

34

units, AB = 6 units)

$A B = (- 5 - 1)^{2} + 0^{2} = 6$
$B C = (- 2 + 5)^{2} + (5 - 0)^{2} = 9 + 25 = 34$
$A C = (- 2 - 1)^{2} + (5 - 0)^{2} = 9 + 25 = 34$
BC = AC, so $△ A B C$ is isosceles.
(Also $A C^{2} + B C^{2} = 68 \neq = 36 = A B^{2}$ , so it is not right-angled.)

Q252 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Standard 30/2/1

In what ratio is the line segment joining the points $(3, - 5)$ and $(- 1, 6)$ divided by the line $y = x$ ?

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Answer: 8 : 7

Let the point dividing the segment in the ratio $k : 1$ be $(\frac{- k + 3}{k + 1}, \frac{6 k - 5}{k + 1})$ .
It lies on $y = x$ , so $- k + 3 = 6 k - 5$ .
$7 k = 8$ , so $k = \frac{8}{7}$ .
The ratio is 8 : 7.

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

Q25 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Standard 30/2/1

A(3, 0), B(6, 4) and C( $- 1$ , 3) are vertices of a triangle ABC. Find length of its median BE.

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Answer: BE

= \frac{5 5}{2}

units

E is the mid-point of AC: $E = (\frac{3 - 1}{2}, \frac{0 + 3}{2}) = (1, \frac{3}{2})$
$B E = (6 - 1)^{2} + (4 - \frac{3}{2})^{2} = 25 + \frac{25}{4} = \frac{125}{4} = \frac{5 5}{2}$ units

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

Q242 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Standard 30/2/2

The vertices of a $△ A B C$ are A( $- 2$ , 4), B(4, 3) and C(1, $- 6$ ). Find length of the median BD.

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Answer: BD

= \frac{145}{2}

units (about 6.02 units)

D is the mid-point of AC: $D = (\frac{- 2 + 1}{2}, \frac{4 - 6}{2}) = (- \frac{1}{2}, - 1)$
$B D = (4 + \frac{1}{2})^{2} + (3 + 1)^{2} = \frac{81}{4} + 16 = \frac{145}{4} = \frac{145}{2}$ units

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Standard 30/2/3

In the given figure, a circle centred at origin O has radius 7 cm, OC is median of $△ O A B$ . Find the length of median OC.

Diagram for CBSE 2024 Class 10 Maths question 23

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Answer: OC

= \frac{7 2}{2}

cm (about 4.95 cm)

From the figure, A lies on the negative y-axis and B on the positive x-axis, on the circle of radius 7: A(0, $- 7$ ), B(7, 0).
C is the mid-point of AB: $C = (\frac{7}{2}, - \frac{7}{2})$ .
$O C = (\frac{7}{2})^{2} + (\frac{7}{2})^{2} = \frac{7}{2} 2 = \frac{7 2}{2}$ cm $\approx 4.95$ cm

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Standard 30/3/1

Find a relation between $x$ and y such that the point P( $x$ , y) is equidistant from the points A(7, 1) and B(3, 5).

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Answer:

x - y = 2

PA = PB, so $P A^{2} = P B^{2}$ .
$(x - 7)^{2} + (y - 1)^{2} = (x - 3)^{2} + (y - 5)^{2}$ .
$x^{2} - 14 x + 49 + y^{2} - 2 y + 1 = x^{2} - 6 x + 9 + y^{2} - 10 y + 25$ .
$- 14 x - 2 y + 50 = - 6 x - 10 y + 34 \Rightarrow - 8 x + 8 y + 16 = 0$ .
So $x - y = 2$ .

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

Q23 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Standard 30/3/1

Points A( $- 1$ , y) and B(5, 7) lie on a circle with centre O(2, $- 3 y$ ) such that AB is a diameter of the circle. Find the value of y. Also, find the radius of the circle.

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Answer:

y = - 1

; radius = 5 units

O is the mid-point of diameter AB.
Mid-point of AB = $(\frac{- 1 + 5}{2}, \frac{y + 7}{2}) = (2, \frac{y + 7}{2})$ .
So $\frac{y + 7}{2} = - 3 y \Rightarrow y + 7 = - 6 y \Rightarrow y = - 1$ .
Then O = (2, 3) and A = ( $- 1$ , $- 1$ ).
Radius OA = $(2 + 1)^{2} + (3 + 1)^{2} = 9 + 16 = 5$ units.

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

Q242 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Standard 30/5/1

Find the ratio in which the point P(–4, 6) divides the line segment joining the points A(–6, 10) and B(3, –8).

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Answer: 2 : 7

Let P divide AB in the ratio $k : 1$ .
x-coordinate: $\frac{3 k - 6}{k + 1} = - 4 \Rightarrow 3 k - 6 = - 4 k - 4 \Rightarrow k = \frac{2}{7}$ .
Check y: $\frac{- 8 \times \frac{2}{7} + 10}{\frac{2}{7} + 1} = \frac{54/7}{9/7} = 6$ . ✓
So P divides AB in the ratio 2 : 7.

Also asked in: 2024 Standard 30/5/2, 2024 Standard 30/5/3

Q24 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2024 · Standard 30/5/1

Prove that the points (3, 0), (6, 4) and (–1, 3) are the vertices of an isosceles triangle.

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Answer: Proved.

Let A(3, 0), B(6, 4), C(–1, 3).
$A B = 3^{2} + 4^{2} = 5$ , $A C = (- 4)^{2} + 3^{2} = 5$ , $B C = (- 7)^{2} + (- 1)^{2} = 50 = 52$ .
AB = AC (and the points are not collinear since $A B + A C = 10 \neq = 52$ ), so ABC is an isosceles triangle.

Also asked in: 2024 Standard 30/5/2, 2024 Standard 30/5/3

Q212 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Basic 430/1/1

Find the coordinates of the point which divides the line segment joining the points $(7, - 1)$ and $(- 3, 4)$ internally in the ratio 2 : 3.

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Answer: (3, 1)

Using the section formula with $m : n = 2 : 3$ :
$x = \frac{2 ( - 3 ) + 3 ( 7 )}{2 + 3} = \frac{- 6 + 21}{5} = 3$
$y = \frac{2 ( 4 ) + 3 ( - 1 )}{2 + 3} = \frac{8 - 3}{5} = 1$
The point is (3, 1).

Also asked in: 2023 Basic 430/1/2, 2023 Basic 430/1/3

Q21 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Basic 430/1/1

Find the value(s) of y for which the distance between the points A(3, $- 1$ ) and B(11, y) is 10 units.

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Answer:

y = 5

y = - 7

$A B^{2} = (11 - 3)^{2} + (y + 1)^{2} = 100$
$64 + (y + 1)^{2} = 100$ , so $(y + 1)^{2} = 36$
$y + 1 = \pm 6$
$y = 5$ or $y = - 7$

Also asked in: 2023 Basic 430/1/2, 2023 Basic 430/1/3

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Basic 430/2/1

Find the value(s) of ‘x’ so that PQ = QR, where the coordinates of P, Q and R are $(6, - 1)$ , $(1, 3)$ and $(x, 8)$ respectively.

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Answer: x = 5 or x =

- 3

$P Q^{2} = (1 - 6)^{2} + (3 + 1)^{2} = 25 + 16 = 41$
$Q R^{2} = (x - 1)^{2} + (8 - 3)^{2} = (x - 1)^{2} + 25$
PQ = QR gives $(x - 1)^{2} + 25 = 41$ , so $(x - 1)^{2} = 16$
$x - 1 = \pm 4$ , so x = 5 or x = $- 3$

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

Q23 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Basic 430/2/1

The vertices of a triangle are $(- 2, 0)$ , $(2, 3)$ and $(1, - 3)$ . Is the triangle equilateral, isosceles or scalene ?

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Answer: Scalene

Let A(−2, 0), B(2, 3), C(1, −3).
$A B = 4^{2} + 3^{2} = 5$
$B C = (- 1)^{2} + (- 6)^{2} = 37$
$C A = (- 3)^{2} + 3^{2} = 18 = 32$
All three sides are different, so the triangle is scalene.

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

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Basic 430/4/1

Find the coordinates of the point which divides the join of A $(- 1, 7)$ and B $(4, - 3)$ in the ratio 2 : 3.

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Answer:

(1, 3)

By the section formula with $m : n = 2 : 3$ :
$x = \frac{2 ( 4 ) + 3 ( - 1 )}{5} = \frac{5}{5} = 1$ .
$y = \frac{2 ( - 3 ) + 3 ( 7 )}{5} = \frac{15}{5} = 3$ .
The point is $(1, 3)$ .

Also asked in: 2023 Basic 430/4/2, 2023 Basic 430/4/3

Q22 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Basic 430/4/1

If the points A $(2, 3)$ , B $(- 5, 6)$ , C $(6, 7)$ and D $(p, 4)$ are the vertices of a parallelogram ABCD, find the value of $p$ .

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Answer:

p = 13

Diagonals of a parallelogram bisect each other, so the mid-points of AC and BD coincide.
Mid-point of AC $= (\frac{2 + 6}{2}, \frac{3 + 7}{2}) = (4, 5)$ .
Mid-point of BD $= (\frac{- 5 + p}{2}, \frac{6 + 4}{2}) = (\frac{p - 5}{2}, 5)$ .
$\frac{p - 5}{2} = 4 \Rightarrow p = 13$ .

Also asked in: 2023 Basic 430/4/2, 2023 Basic 430/4/3

Q242 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Basic 430/5/1

Show that A(1, 2), B(5, 4), C(3, 8) and D( $-$ 1, 6) are vertices of a parallelogram ABCD.

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Answer: Proved.

AB $= 4^{2} + 2^{2} = 20$ , CD $= (- 4)^{2} + (- 2)^{2} = 20$ .
BC $= (- 2)^{2} + 4^{2} = 20$ , DA $= 2^{2} + (- 4)^{2} = 20$ .
Both pairs of opposite sides are equal (AB = CD and BC = DA), so ABCD is a parallelogram.
(Check: midpoint of AC = (2, 5) = midpoint of BD, so the diagonals bisect each other.)

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

Q24 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Basic 430/5/1

Show that the points A(3, 0), B(6, 4) and C( $-$ 1, 3) are vertices of a right-angled triangle.

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Answer: Proved.

$A B^{2} = (6 - 3)^{2} + (4 - 0)^{2} = 25$ .
$A C^{2} = (- 1 - 3)^{2} + (3 - 0)^{2} = 25$ .
$B C^{2} = (- 1 - 6)^{2} + (3 - 4)^{2} = 50$ .
$A B^{2} + A C^{2} = 50 = B C^{2}$ , so by the converse of Pythagoras theorem the triangle is right-angled at A.

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

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Standard 30/2/1

Show that the points $(- 2, 3)$ , $(8, 3)$ and $(6, 7)$ are the vertices of a right-angled triangle.

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Answer: Proved.

Let A $(- 2, 3)$ , B $(8, 3)$ , C $(6, 7)$ .
$A B^{2} = 1 0^{2} + 0^{2} = 100$
$B C^{2} = (- 2)^{2} + 4^{2} = 20$
$A C^{2} = 8^{2} + 4^{2} = 80$
$B C^{2} + A C^{2} = 20 + 80 = 100 = A B^{2}$ , so by the converse of Pythagoras theorem the triangle is right-angled at C.

Also asked in: 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q212 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Standard 30/5/1

The line segment joining the points $A (4, - 5)$ and $B (4, 5)$ is divided by the point P such that AP : AB = 2 : 5. Find the coordinates of P.

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Answer: P(4,

- 1

)

AP : AB = 2 : 5, so AP : PB = 2 : 3.
$x = \frac{2 \times 4 + 3 \times 4}{5} = 4$ .
$y = \frac{2 \times 5 + 3 \times ( - 5 )}{5} = \frac{- 5}{5} = - 1$ .
P is (4, $- 1$ ).

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q21 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Standard 30/5/1

Point P(x, y) is equidistant from points A(5, 1) and B(1, 5). Prove that x = y.

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Answer: Proved.

PA = PB, so $P A^{2} = P B^{2}$ .
$(x - 5)^{2} + (y - 1)^{2} = (x - 1)^{2} + (y - 5)^{2}$ .
$x^{2} - 10 x + 25 + y^{2} - 2 y + 1 = x^{2} - 2 x + 1 + y^{2} - 10 y + 25$ .
$- 10 x - 2 y = - 2 x - 10 y$ , so $- 8 x = - 8 y$ .
Hence x = y.

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q242 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Standard 30/5/1

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

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Answer: 5 : 1 (point of division

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

)

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

Also asked in: 2023 Standard 30/5/3

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Standard 30/5/2

Find the ratio in which line $y = x$ divides the line segment joining the points $(6, - 3)$ and $(1, 6)$ .

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Answer: 9 : 5

Let the line divide the segment in the ratio k : 1.
Point of division $= (\frac{k + 6}{k + 1}, \frac{6 k - 3}{k + 1})$ .
It lies on $y = x$ : $6 k - 3 = k + 6$ , so $5 k = 9$ , $k = \frac{9}{5}$ .
Ratio = 9 : 5 (point of division $(\frac{39}{14}, \frac{39}{14})$ ).

Q222 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Standard 30/6/1

A line intersects y-axis and x-axis at point P and Q, respectively. If R(2, 5) is the mid-point of line segment PQ, then find the coordinates of P and Q.

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Answer: P(0, 10) and Q(4, 0)

Let P = (0, b) on the y-axis and Q = (a, 0) on the x-axis.
Mid-point of PQ = $(\frac{a}{2}, \frac{b}{2}) = (2, 5)$ .
So a = 4 and b = 10.
P = (0, 10), Q = (4, 0).

Q242 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Standard 30/6/2

Find the points on the $x$ -axis, each of which is at a distance of 10 units from the point A(11, – 8).

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Answer: (17, 0) and (5, 0)

Let the point be P(x, 0).
$P A^{2} = (x - 11)^{2} + (0 + 8)^{2} = 100$
$(x - 11)^{2} = 36$ , so $x - 11 = \pm 6$ .
x = 17 or x = 5. The points are (17, 0) and (5, 0).

Q212 marksVery Short AnswerCoordinate Geometry CBSE 2023 · Standard 30/6/3

Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4).

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Answer: 5 : 1

Let the y-axis divide the segment in the ratio k : 1 at (0, y).
x-coordinate: $\frac{k ( - 1 ) + 1 ( 5 )}{k + 1} = 0$
$- k + 5 = 0$ , so k = 5.
Ratio = 5 : 1

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