Chapter 7 - Maths - Question Paper

CBSE Class 10 — Coordinate Geometry — 80 Marks

Central Board of Secondary Education

MATHEMATICS (Standard) — Code 041

Chapter 7: Coordinate Geometry — Comprehensive Practice Paper

Class: X Time: 3 Hours

Maximum Marks: 80

Roll No: ___________ Date: ___________

General Instructions

1. This question paper contains 38 questions divided into four sections A, B, C, and D.
2. Section A contains 20 questions of 1 mark each (Q1–Q18 MCQ; Q19–Q20 Assertion-Reason).
3. Section B contains 5 questions of 2 marks each (Q21–Q25).
4. Section C contains 6 questions of 3 marks each (Q26–Q31).
5. Section D contains 4 questions of 5 marks each (Q32–Q35), with internal choice in Q34 and Q35.
6. Section E contains 3 Case-Based Questions of 4 marks each (Q36–Q38).
7. All questions are compulsory. However, an internal choice is provided in some questions.
8. Use of calculators is not permitted.

A

Section A — Multiple Choice Questions

Each question carries 1 mark. Choose the most appropriate option.

20 × 1 = 20 Marks

1.

The distance between the points \(A(2, 3)\) and \(B(4, 1)\) is:

(a) \(2\sqrt{2}\)

(b) \(2\sqrt{3}\)

(c) \(4\)

(d) \(\sqrt{6}\)

1

2.

The distance of point \(P(8, 15)\) from the origin is:

(a) \(23\)

(b) \(17\)

(c) \(\sqrt{169}\)

(d) \(15\)

1

3.

The midpoint of the segment joining \((-1, 7)\) and \((3, -3)\) is:

(a) \((1, 2)\)

(b) \((2, 4)\)

(c) \((0, 2)\)

(d) \((1, 4)\)

1

4.

The point which divides the segment joining \(A(1, 3)\) and \(B(3, 9)\) in ratio \(1:2\) internally is:

(a) \((2, 6)\)

(b) \(\left(\dfrac{5}{3}, 5\right)\)

(c) \((1, 3)\)

(d) \((3, 9)\)

1

5.

The distance between \((a, b)\) and \((-a, -b)\) is:

(a) \(\sqrt{a^2+b^2}\)

(b) \(2(a+b)\)

(c) \(2\sqrt{a^2+b^2}\)

(d) \(4\sqrt{a^2+b^2}\)

1

6.

If the distance between \(P(2, -3)\) and \(Q(10, y)\) is 10 units, the value(s) of \(y\) are:

(a) \(3\) only

(b) \(-9\) only

(c) \(3\) and \(-9\)

(d) \(9\) and \(-3\)

1

7.

The point on the \(x\)-axis equidistant from \((2, -5)\) and \((-2, 9)\) is:

(a) \((7, 0)\)

(b) \((-7, 0)\)

(c) \((0, -7)\)

(d) \((0, 7)\)

1

8.

The ratio in which the point \((-4, 6)\) divides the segment joining \((-6, 10)\) and \((3, -8)\) is:

(a) \(3:7\)

(b) \(7:2\)

(c) \(2:7\)

(d) \(1:2\)

1

9.

The \(y\)-axis divides the segment joining \((5, -6)\) and \((-1, -4)\) in ratio:

(a) \(1:5\)

(b) \(5:1\)

(c) \(1:1\)

(d) \(2:3\)

1

10.

\((1, 7)\), \((4, 2)\), \((-1, -1)\), \((-4, 4)\) are the vertices of a:

(a) Rectangle

(b) Rhombus

(c) Square

(d) Parallelogram only

1

11.

In parallelogram \(ABCD\), \(A(6,1)\), \(B(8,2)\), \(C(9,4)\), \(D(p,3)\). The value of \(p\) is:

(a) \(5\)

(b) \(6\)

(c) \(7\)

(d) \(8\)

1

12.

Area of a rhombus with diagonals \(d_1 = 4\sqrt{2}\) and \(d_2 = 6\sqrt{2}\) is:

(a) \(20\) sq units

(b) \(24\) sq units

(c) \(48\) sq units

(d) \(12\) sq units

1

13.

The points \((1, 5)\), \((2, 3)\), and \((-2, -11)\) are:

(a) Collinear

(b) Vertices of an equilateral triangle

(c) Not collinear

(d) Vertices of a right triangle

1

14.

If \((1, 2)\), \((4, y)\), \((x, 6)\), \((3, 5)\) are vertices of a parallelogram (in order), then \(x + y\) equals:

(a) \(5\)

(b) \(6\)

(c) \(8\)

(d) \(11\)

1

15.

The coordinates of the point \(A\), where \(AB\) is the diameter of a circle with centre \((2, -3)\) and \(B(1, 4)\), are:

(a) \((3, -10)\)

(b) \((3, -8)\)

(c) \((-3, 10)\)

(d) \((1, -4)\)

1

16.

If point \(P(k, 0)\) divides the segment joining \(A(2, -2)\) and \(B(-7, 4)\) in ratio \(1:2\), then \(k\) equals:

(a) \(-1\)

(b) \(1\)

(c) \(-\dfrac{1}{3}\)

(d) \(\dfrac{1}{3}\)

1

17.

What is the distance between the two trisection points of the segment joining \(A(2, -2)\) and \(B(-7, 4)\)?

(a) \(\sqrt{117}/3\)

(b) \(\sqrt{117}\)/3

(c) \(\sqrt{13}\)

(d) \(3\sqrt{13}/2\)

1

18.

A point \(P\) on segment \(AB\) is such that \(AP = \dfrac{3}{7}AB\), where \(A(-2,-2)\) and \(B(2,-4)\). The coordinates of \(P\) are:

(a) \(\left(\dfrac{-2}{7}, \dfrac{-20}{7}\right)\)

(b) \(\left(\dfrac{-8}{7}, \dfrac{-20}{7}\right)\)

(c) \(\left(\dfrac{2}{7}, \dfrac{-22}{7}\right)\)

(d) \(\left(\dfrac{-2}{7}, \dfrac{-22}{7}\right)\)

1

19.

Assertion (A): The point \((-7, 0)\) lies on the \(x\)-axis and is equidistant from \((2, -5)\) and \((-2, 9)\).

Reason (R): Any point equidistant from two given points lies on the perpendicular bisector of the segment joining them.

(a) Both A and R are true; R is the correct explanation of A

(b) Both A and R are true; R is not the correct explanation of A

(c) A is true but R is false

(d) A is false but R is true

1

20.

Assertion (A): The diagonals of a parallelogram bisect each other, so their midpoints are equal.

Reason (R): The midpoint formula is used to verify that the diagonals of a parallelogram are equal in length.

(a) Both A and R are true; R is the correct explanation of A

(b) Both A and R are true; R is not the correct explanation of A

(c) A is true but R is false

(d) A is false but R is true

1

B

Section B — Short Answer I

Each question carries 2 marks. Show all working.

5 × 2 = 10 Marks

21.

Find the value(s) of \(y\) if the distance between points \(P(2, -3)\) and \(Q(10, y)\) is 10 units.

2

22.

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

2

23.

Determine if the points \((1, 5)\), \((2, 3)\), and \((-2, -11)\) are collinear.

2

24.

Find the ratio in which the line segment joining \(A(1, -5)\) and \(B(-4, 5)\) is divided by the \(x\)-axis. Also find the coordinates of the point of division.

2

25.

If \((1, 2)\), \((4, y)\), \((x, 6)\), and \((3, 5)\) are the vertices of a parallelogram taken in order, find \(x\) and \(y\).

2

C

Section C — Short Answer II

Each question carries 3 marks. Show all working clearly.

6 × 3 = 18 Marks

26.

Find the point on the \(x\)-axis which is equidistant from \((2, -5)\) and \((-2, 9)\).

3

27.

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

3

28.

Find the ratio in which the segment joining \((-3, 10)\) and \((6, -8)\) is divided by \((-1, 6)\). Also verify using the \(y\)-coordinate.

3

29.

Check whether \((5, -2)\), \((6, 4)\), and \((7, -2)\) are vertices of an isosceles triangle. Also find the area of the triangle.

3

30.

Find the coordinates of point \(A\), where \(AB\) is the diameter of a circle whose centre is \((2, -3)\) and \(B(1, 4)\).

3

31.

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

3

D

Section D — Long Answer

Each question carries 5 marks. Internal choice is provided in Q34 and Q35.

4 × 5 = 20 Marks

32.

Show that the points \((1, 7)\), \((4, 2)\), \((-1, -1)\), and \((-4, 4)\) are the vertices of a square. Find the side length, diagonal length, and verify that diagonal\(^2 = 2 \times \)side\(^2\).

5

33.

Find the area of a rhombus whose vertices are \((3, 0)\), \((4, 5)\), \((-1, 4)\), and \((-2, -1)\) taken in order. Also verify that the diagonals bisect each other by showing their midpoints are equal.

5

34.

If A and B are \((-2, -2)\) and \((2, -4)\) respectively, find the coordinates of P such that \(AP = \dfrac{3}{7}AB\) and P lies on segment AB.

OR

Find the coordinates of the vertices of a triangle whose midpoints are \((1, 2)\), \((0, -1)\), and \((2, -1)\). [Hint: use three simultaneous midpoint equations.]

5

35.

If \(Q(0, 1)\) is equidistant from \(P(5, -3)\) and \(R(x, 6)\), find the values of \(x\). Also find the distances \(QR\) and \(PR\) for each value of \(x\).

OR

Find a relation between \(x\) and \(y\) such that the point \((x, y)\) is equidistant from \((3, 6)\) and \((-3, 4)\). What geometric figure does this relation represent?

5

E

Section E — Case-Based Questions

Each question carries 4 marks. Read each situation carefully and answer all parts.

3 × 4 = 12 Marks

36.

Relay Tower Problem. A telephone company wants to position a relay tower at point P between towns A and B such that the distance of the tower from B is twice its distance from A. Taking A as the origin and 1 km = 1 unit, the coordinates of B are \((36, 15)\). The tower must lie on segment AB.

(i) In what ratio does P divide AB?  [1 mark]

(ii) Find the coordinates of P.  [1 mark]

(iii) Find the distance AP and verify it equals half of BP.  [2 marks]

37.

Sports Day Setup. In a rectangular school ground ABCD, lines are drawn with chalk powder at 1 m intervals. 100 flower pots are placed at 1 m intervals along AD. Niharika runs \(\dfrac{1}{4}\)th of AD on the 2nd line and posts a green flag at G. Preet runs \(\dfrac{1}{5}\)th of AD on the 8th line and posts a red flag at R. AD = 100 m.

(i) What are the coordinates of G and R?  [1 mark]

(ii) Find the distance GR between the two flags.  [1 mark]

(iii) Rashmi must post a blue flag exactly halfway between G and R. Find where she should post it (line number and distance along AD).  [2 marks]

38.

Seismology Application. A seismological station at origin O detects an earthquake. Sensor A is placed at \((6, 0)\) km and sensor B at \((0, 8)\) km from O. The epicentre E lies on the perpendicular bisector of segment AB and is equidistant from A and B.

(i) Find the midpoint M of segment AB.  [1 mark]

(ii) Find the length AB.  [1 mark]

(iii) If the epicentre lies on the \(y\)-axis (i.e., \(E = (0, y)\)), find the coordinates of E and its distances from A and B.  [2 marks]

Section	No. of Questions	Marks Each	Total Marks
A — MCQ + Assertion-Reason	20	1	20
B — Short Answer I	5	2	10
C — Short Answer II	6	3	18
D — Long Answer	4	5	20
E — Case Based	3	4	12
Grand Total	38	—	80

*** END OF QUESTION PAPER ***   •   CBSE Class X — Mathematics (Standard) Code 041   •   Chapter 7: Coordinate Geometry

🗝 Detailed Answer Key

Full step-by-step solutions with marking scheme — for teacher / self-evaluation use

Section A — MCQ Answers (1 mark each)

Q1 1 mark

Ans: (a) \(d = \sqrt{(4-2)^2+(1-3)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}\)

Q2 1 mark

Ans: (b) \(OP = \sqrt{8^2+15^2} = \sqrt{64+225} = \sqrt{289} = 17\). Note: \((8, 15, 17)\) is a Pythagorean triple.

Q3 1 mark

Ans: (a) \(M = \left(\dfrac{-1+3}{2}, \dfrac{7+(-3)}{2}\right) = (1, 2)\)

Q4 1 mark

Ans: (b) \(x = \dfrac{1(3)+2(1)}{3} = \dfrac{5}{3}\),   \(y = \dfrac{1(9)+2(3)}{3} = 5\).   Point: \(\left(\dfrac{5}{3}, 5\right)\)

Q5 1 mark

Ans: (c) \(d = \sqrt{(-2a)^2+(-2b)^2} = \sqrt{4a^2+4b^2} = 2\sqrt{a^2+b^2}\)

Q6 1 mark

Ans: (c) \(\sqrt{64+(y+3)^2}=10 \Rightarrow (y+3)^2=36 \Rightarrow y+3=\pm 6 \Rightarrow y=3\) or \(y=-9\)

Q7 1 mark

Ans: (b) Let \(P(x,0)\). Setting \(PA^2=PB^2\): \((x-2)^2+25=(x+2)^2+81 \Rightarrow -8x=56 \Rightarrow x=-7\). Point: \((-7, 0)\).

Q8 1 mark

Ans: (c) Let ratio \(k:1\). \(\dfrac{3k-6}{k+1}=-4 \Rightarrow 7k=2 \Rightarrow k=\dfrac{2}{7}\). Ratio \(= 2:7\).

Q9 1 mark

Ans: (b) On \(y\)-axis, \(x=0\). Let ratio \(k:1\): \(\dfrac{-k+5}{k+1}=0 \Rightarrow k=5\). Ratio \(= 5:1\).

Q10 1 mark

Ans: (c) All four sides \(=\sqrt{34}\) and both diagonals \(=\sqrt{68}\). Since sides equal and diagonals equal \(\Rightarrow\) Square.

Q11 1 mark

Ans: (c) Mid(AC) = Mid(BD): \(\left(\dfrac{15}{2},\dfrac{5}{2}\right)=\left(\dfrac{8+p}{2},\dfrac{5}{2}\right) \Rightarrow p=7\).

Q12 1 mark

Ans: (b) Area \(= \dfrac{1}{2}\times d_1 \times d_2 = \dfrac{1}{2}\times 4\sqrt{2}\times 6\sqrt{2} = \dfrac{1}{2}\times 48 = 24\) sq units.

Q13 1 mark

Ans: (c) \(AB=\sqrt{5}\), \(BC=2\sqrt{53}\), \(AC=\sqrt{265}\). Since \(AB+BC \neq AC\), points are not collinear.

Q14 1 mark

Ans: (c) Mid(AC) = Mid(BD): \(x=6\), \(y=2\). So \(x+y=8\).

Q15 1 mark

Ans: (a) Centre is midpoint of AB: \(\left(\dfrac{x_A+1}{2},\dfrac{y_A+4}{2}\right)=(2,-3)\). So \(x_A=3\), \(y_A=-10\). \(A=(3,-10)\).

Q16 1 mark

Ans: (a) Ratio \(1:2\): \(x=\dfrac{1(-7)+2(2)}{3}=\dfrac{-3}{3}=-1\). So \(k=-1\). (Verify: \(y=\dfrac{1(4)+2(-2)}{3}=0\) ✓)

Q17 1 mark

Ans: (c) Trisection points P\((-1,0)\) and Q\((-4,2)\). Distance \(PQ=\sqrt{9+4}=\sqrt{13}\).

Q18 1 mark

Ans: (d) \(AP=\dfrac{3}{7}AB \Rightarrow\) ratio \(AP:PB=3:4\). \(P_x=\dfrac{3(2)+4(-2)}{7}=\dfrac{-2}{7}\), \(P_y=\dfrac{3(-4)+4(-2)}{7}=\dfrac{-20}{7}.\) Wait — \(P_y = \dfrac{3(-4)+4(-2)}{7}=\dfrac{-20}{7}\). So \(P=\left(-\dfrac{2}{7},-\dfrac{20}{7}\right)\).

Q19 — Assertion-Reason 1 mark

Ans: (a) Both A and R are true; R correctly explains A. The point \((-7,0)\) lies on the \(x\)-axis (so \(y=0\)) and is on the perpendicular bisector of the segment joining \((2,-5)\) and \((-2,9)\), making it equidistant from both.

Q20 — Assertion-Reason 1 mark

Ans: (c) A is true but R is false. The midpoint formula is used to show diagonals bisect each other (same midpoint), not that they are equal in length. Equal diagonals would make it a rectangle, not every parallelogram.

Section B — Short Answer I (2 marks each)

Q21 2 marks

Set up

\(\sqrt{(10-2)^2+(y+3)^2}=10 \Rightarrow 64+(y+3)^2=100\)

Solve

\((y+3)^2=36 \Rightarrow y+3=\pm 6 \Rightarrow y=3\) or \(y=-9\)

✓ y = 3 or y = −9  [1 mark per value]

Q22 2 marks

Formula

\(m_1=2, m_2=3\), \(A(-1,7)\), \(B(4,-3)\)

x

\(x=\dfrac{2(4)+3(-1)}{5}=\dfrac{5}{5}=1\)

y

\(y=\dfrac{2(-3)+3(7)}{5}=\dfrac{15}{5}=3\)

✓ Point = (1, 3)  [1 mark formula + 1 mark answer]

Q23 2 marks

Distances

\(AB=\sqrt{1+4}=\sqrt{5}\),   \(BC=\sqrt{16+196}=\sqrt{212}=2\sqrt{53}\),   \(AC=\sqrt{9+256}=\sqrt{265}\)

Check

\(AB+BC \approx 2.24+14.56=16.80 \neq \sqrt{265}\approx 16.28\). So \(AB+BC\neq AC\).

✓ Points are NOT collinear.  [1 + 1 marks]

Q24 2 marks

Ratio

On \(x\)-axis, \(y=0\). Let ratio \(k:1\): \(\dfrac{5k+(-5)}{k+1}=0 \Rightarrow 5k-5=0 \Rightarrow k=1\). Ratio \(= 1:1\).

Point

\(x=\dfrac{1(-4)+1(1)}{2}=\dfrac{-3}{2}\). Division point \(=\left(-\dfrac{3}{2},0\right)\).

✓ Ratio = 1:1; Point = (−3/2, 0)  [1 + 1 marks]

Q25 2 marks

Diagonals

Diagonals are AC and BD. Mid(AC) = Mid(BD) for a parallelogram.

Equate

\(\left(\dfrac{1+x}{2},\dfrac{2+6}{2}\right)=\left(\dfrac{4+3}{2},\dfrac{y+5}{2}\right)\) ⇒ \(x=6\), \(y=2\).

✓ x = 6, y = 2  [1 + 1 marks]

Section C — Short Answer II (3 marks each)

Q26 3 marks

Assume

Let point be \(P(x,0)\) on \(x\)-axis. Need \(PA=PB\), so \(PA^2=PB^2\).

Set up

\((x-2)^2+25=(x+2)^2+81\)

Expand

\(x^2-4x+4+25=x^2+4x+4+81 \Rightarrow -8x=56 \Rightarrow x=-7\)

✓ Required point = (−7, 0)  [1+1+1 marks]

Q27 3 marks

P (ratio 1:2)

\(P_x=\dfrac{1(-2)+2(4)}{3}=2\),   \(P_y=\dfrac{1(-3)+2(-1)}{3}=-\dfrac{5}{3}\)

Q (ratio 2:1)

\(Q_x=\dfrac{2(-2)+1(4)}{3}=0\),   \(Q_y=\dfrac{2(-3)+1(-1)}{3}=-\dfrac{7}{3}\)

✓ P = (2, −5/3) and Q = (0, −7/3)  [1+1+1 marks]

Q28 3 marks

x-coord

Let ratio \(k:1\): \(\dfrac{6k-3}{k+1}=-1 \Rightarrow 6k-3=-k-1 \Rightarrow 7k=2 \Rightarrow k=\dfrac{2}{7}\)

Ratio

Ratio \(= 2:7\)

Verify y

\(y=\dfrac{2(-8)+7(10)}{9}=\dfrac{54}{9}=6\) ✓

✓ Ratio = 2 : 7  [1+1+1 marks]

Q29 3 marks

Sides

\(AB=\sqrt{1+36}=\sqrt{37}\),   \(BC=\sqrt{1+36}=\sqrt{37}\),   \(CA=\sqrt{4+0}=2\)

Isosceles?

\(AB=BC=\sqrt{37}\neq CA=2\) ⇒ Isosceles ✓

Area

Base \(CA=2\). Height = \(y\)-distance from B to line AC: \(|4-(-2)|=6\). Area \(=\dfrac{1}{2}\times 2\times 6=6\) sq units.

✓ Isosceles (AB = BC = √37); Area = 6 sq units  [1+1+1 marks]

Q30 3 marks

Property

Centre is midpoint of diameter AB: \(\left(\dfrac{x_A+1}{2},\dfrac{y_A+4}{2}\right)=(2,-3)\)

Solve x

\(\dfrac{x_A+1}{2}=2 \Rightarrow x_A=3\)

Solve y

\(\dfrac{y_A+4}{2}=-3 \Rightarrow y_A=-10\)

✓ A = (3, −10)  [1+1+1 marks]

Q31 3 marks

Strategy

Three points divide AB into 4 equal parts. Use ratios 1:3, 1:1, 3:1.

P (1:3)

\(P=\left(\dfrac{1(2)+3(-2)}{4},\dfrac{1(8)+3(2)}{4}\right)=\left(-1,\dfrac{14}{4}\right)=\left(-1,\dfrac{7}{2}\right)\)

Q (1:1)

\(Q=\left(\dfrac{-2+2}{2},\dfrac{2+8}{2}\right)=(0,5)\)

R (3:1)

\(R=\left(\dfrac{3(2)+1(-2)}{4},\dfrac{3(8)+1(2)}{4}\right)=\left(1,\dfrac{13}{2}\right)\)

✓ P = (−1, 7/2), Q = (0, 5), R = (1, 13/2)  [1+1+1 marks]

Section D — Long Answer (5 marks each)

Q32 5 marks

Sides

\(AB=\sqrt{(4-1)^2+(2-7)^2}=\sqrt{9+25}=\sqrt{34}\)
\(BC=\sqrt{(-1-4)^2+(-1-2)^2}=\sqrt{25+9}=\sqrt{34}\)
\(CD=\sqrt{(-4+1)^2+(4+1)^2}=\sqrt{9+25}=\sqrt{34}\)
\(DA=\sqrt{(1+4)^2+(7-4)^2}=\sqrt{25+9}=\sqrt{34}\)

Diagonals

\(AC=\sqrt{(-1-1)^2+(-1-7)^2}=\sqrt{4+64}=\sqrt{68}\)
\(BD=\sqrt{(-4-4)^2+(4-2)^2}=\sqrt{64+4}=\sqrt{68}\)

Verify

Diagonal\(^2=68=2\times 34=2\times\)Side\(^2\) ✓ (confirms right angles at vertices)

Conclude

All sides equal (\(\sqrt{34}\)), both diagonals equal (\(\sqrt{68}\)) ⇒ ABCD is a Square.

✓ Side = √34 units, Diagonal = √68 units, Diagonal² = 2 × Side²  [1+1+1+1+1 marks]

Q33 5 marks

Diagonals

Vertices: \(A(3,0)\), \(B(4,5)\), \(C(-1,4)\), \(D(-2,-1)\). Diagonals: AC and BD.

d₁ = AC

\(d_1=\sqrt{(-1-3)^2+(4-0)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\)

d₂ = BD

\(d_2=\sqrt{(-2-4)^2+(-1-5)^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}\)

Area

\(\text{Area}=\dfrac{1}{2}\times 4\sqrt{2}\times 6\sqrt{2}=\dfrac{1}{2}\times 48=24\) sq units

Midpoints

Mid(AC)\(=\left(\dfrac{3-1}{2},\dfrac{0+4}{2}\right)=(1,2)\). Mid(BD)\(=\left(\dfrac{4-2}{2},\dfrac{5-1}{2}\right)=(1,2)\). Equal ✓ ⇒ diagonals bisect each other.

✓ Area = 24 sq units; diagonals bisect each other confirmed  [1+1+1+1+1 marks]

Q34 5 marks

Ratio

\(AP=\dfrac{3}{7}AB\) means \(AP:PB=3:4\). So P divides AB in ratio 3:4.

x

\(P_x=\dfrac{3(2)+4(-2)}{7}=\dfrac{6-8}{7}=\dfrac{-2}{7}\)

y

\(P_y=\dfrac{3(-4)+4(-2)}{7}=\dfrac{-12-8}{7}=\dfrac{-20}{7}\)

Verify

P lies on AB since \(AP:PB=3:4\) and \(AP+PB=AB\) ✓

✓ P = (−2/7, −20/7)  [1+1+1+1+1 marks]

OR: Let vertices be \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\). Midpoints: \(M_1=(1,2)\) of BC, \(M_2=(0,-1)\) of CA, \(M_3=(2,-1)\) of AB. From midpoint equations: \(x_2+x_3=2\), \(x_3+x_1=0\), \(x_1+x_2=4\) ⇒ \(x_1=1\), \(x_2=3\), \(x_3=-1\). Similarly: \(y_1=-4\), \(y_2=0\), \(y_3=4\). Vertices: \(A(1,-4)\), \(B(3,0)\), \(C(-1,4)\).

Q35 5 marks

QP = QR

\(QP^2=(5-0)^2+(-3-1)^2=25+16=41\)

Set equal

\(QR^2=(x-0)^2+(6-1)^2=x^2+25=41 \Rightarrow x^2=16 \Rightarrow x=\pm 4\)

QR

\(QR=\sqrt{41}\) for both \(x=4\) and \(x=-4\)

PR (x=4)

\(PR=\sqrt{(5-4)^2+(-3-6)^2}=\sqrt{1+81}=\sqrt{82}\)

PR (x=−4)

\(PR=\sqrt{(5+4)^2+(-3-6)^2}=\sqrt{81+81}=9\sqrt{2}\)

✓ x = 4 or −4; QR = √41 in both cases; PR = √82 (x=4) or 9√2 (x=−4)  [1+2+1+1 marks]

OR: Let \(P(x,y)\) equidistant from \(A(3,6)\) and \(B(-3,4)\). PA\(^2\)=PB\(^2\):
\((x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2\)
\(x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16\)
\(-6x-12y+36=6x-8y+16 \Rightarrow -12x-4y+20=0 \Rightarrow 3x+y=5\)
This is a straight line (the perpendicular bisector of AB).

Section E — Case-Based Questions (4 marks each)

Q36 — Relay Tower 4 marks

(i) Ratio

Distance from B = twice distance from A ⇒ \(PB=2 \cdot PA\) ⇒ \(PA:PB=1:2\)  [1 mark]

(ii) Coords of P

\(P_x=\dfrac{1(36)+2(0)}{3}=12\),   \(P_y=\dfrac{1(15)+2(0)}{3}=5\). So \(P=(12,5)\).  [1 mark]

(iii) AP

\(AP=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\) km  [1 mark]

Verify

\(BP=\sqrt{(36-12)^2+(15-5)^2}=\sqrt{576+100}=\sqrt{676}=26\) km. Since \(26=2\times 13\), \(BP=2\cdot AP\) ✓  [1 mark]

✓ Ratio 1:2; P = (12,5); AP = 13 km; BP = 26 km = 2 × AP

Q37 — Sports Day 4 marks

(i) G and R

G: 2nd line, \(\dfrac{1}{4}\times 100=25\)m along AD ⇒ \(G=(2,25)\). R: 8th line, \(\dfrac{1}{5}\times 100=20\)m along AD ⇒ \(R=(8,20)\).  [1 mark]

(ii) GR

\(GR=\sqrt{(8-2)^2+(20-25)^2}=\sqrt{36+25}=\sqrt{61}\approx 7.81\) m  [1 mark]

(iii) Blue flag

Midpoint of GR: \(\left(\dfrac{2+8}{2},\dfrac{25+20}{2}\right)=\left(5,22.5\right)\)  [1 mark]

Interpret

Rashmi posts blue flag on the 5th line, \(22.5\) m along AD from A.  [1 mark]

✓ G=(2,25); R=(8,20); GR=√61 m; Blue flag at 5th line, 22.5 m from A

Q38 — Seismology 4 marks

(i) Midpoint M

\(M=\left(\dfrac{6+0}{2},\dfrac{0+8}{2}\right)=(3,4)\)  [1 mark]

(ii) Length AB

\(AB=\sqrt{(6-0)^2+(0-8)^2}=\sqrt{36+64}=\sqrt{100}=10\) km  [1 mark]

(iii) E on y-axis

Let \(E=(0,y)\). \(EA=EB\) so \(EA^2=EB^2\):
\((0-6)^2+(y-0)^2=(0-0)^2+(y-8)^2\)
\(36+y^2=y^2-16y+64 \Rightarrow 16y=28 \Rightarrow y=\dfrac{7}{4}\)  [1 mark]

Distances

\(EA=\sqrt{36+\left(\dfrac{7}{4}\right)^2}=\sqrt{36+\dfrac{49}{16}}=\sqrt{\dfrac{625}{16}}=\dfrac{25}{4}\) km. \(EB=\dfrac{25}{4}\) km ✓  [1 mark]

✓ M=(3,4); AB=10 km; E=(0, 7/4); EA=EB=25/4 km

*** END OF ANSWER KEY ***   •   All solutions follow CBSE marking scheme   •   Chapter 7: Coordinate Geometry