Contents

**GSEB Solutions for Class 10 mathematics – Coordinate Geometry (English Medium)**

### Exercise-8.1

**Question 1:**

Find the distance between the following pairs of points :

- (2, -3), (7, 9)
- (-3, 4), (0, 0)
- (a + b, a – b), (b – a, a + b)

**Solution :**

**Question 2:**

A point P on X-axis is at distance 13 from A(11, 12). Find the coordinates of P.

**Solution :**

The y-coordinate of any point on the x-axis is zero.

Let, P(x, 0) be a point on the x-axis such that the distance of it from A(11, 12) is 13 units.

PA = 13

∴ PA^{2 }= 169

∴ (x – 11)^{2 }+ (0 – 12)^{2 }= 169

∴ x^{2 }– 22x + 121 + 144 = 169

∴ x^{2 }– 22x + 265 = 169

∴ x^{2 }– 22x + 96 = 0

∴ x^{2 }– 16x – 6x + 96 = 0

∴ (x – 16)(x – 6) = 0

∴ x – 16 = 0 or x – 6 = 0

∴ x = 16 or x = 6

∴ The required point on x-axis is P(x, 0) = P(16, 0) or P(x, 0) = P(6, 0).

**Question 3:**

Using distance formula, show that A(2, 3), B(4, 7) and C(0, -1) are collinear points.

**Solution :**

**Question 4:**

Find a point on Y-axis which is equidistant from P(-6, 4) and Q(2, -8).

**Solution :**

**Question 5:**

P(x, y) is a point on the perpendicular bisector of the segment \(\overline{AB}\) joining A(2, 3) and B(-4, 1). Find the relation between x and y.

**Solution :**

**Question 6:**

If the distance between A(10, 8) and B(a, 2) is 10, find a.

**Solution :**

Here, AB =10

AB = 100

∴ (10 – a)^{2 }+ (8 – 2)^{2 }= 100

∴ 100 – 20a + a^{2 }+ 36 = 100

∴ a^{2 }– 20a + 36 = 0

∴ a(a – 18)(a – 2) = 0

∴ (a – 18)(a – 2) = 0

∴ a – 18 = 0 or a – 2 = 0

∴ a = 18 or a = 2

So if the distance between A(10, 8) and B(a, 2) is 10, then a = 18 or a = 2.

**Question 7:**

m∠B = 90 in the triangle whose vertices are A(2, 3), B(4, 5) and C(a, 2). Find a.

**Solution :**

**Question 8:**

A(-3, O) and B(3, O) are the vertices of an equilateral ∆ABC. Find coordinates of C.

**Solution :**

**Question 9:**

A(1, 7), B(2, 4), C(k, 5) are the vertices of a right angled ∆ABC,

- Find k, if ∠A is a right angle
- Find k, if ∠B is a right angle
- Find k, if ∠C is a right angle

**Solution :**

Here, A(1, 7), B(2, 4), C(k, 5) are the vertices of a right angled triangle.

AB^{2 }= (1 – 2)^{2 }+ (7 – 4)^{2 }= 1 + 9 = 10

BC^{2 }= (2 – k)^{2 }+ (4 – 5)^{2}

= 4 – 4k + k^{2 }+ 1

= k^{2 }– 4k + 5

AC^{2 }= (1 – k)^{2 }+ (7 – 5)^{2}

= 1 – 2k + k^{2 }+ 4

= k^{2 }– 2k + 5

1. ∠A is a right angle.

∴ BC^{2 }= AB^{2 }+AC^{2}

∴ k^{2 }– 4k + 5 = 10 + k^{2 }– 2k + 5

∴ -2 k = 10

∴ k = -5

Hence, if ∠A is a right angle in ∆ABC, then k = -5.

2. ∠B is a right angle.

∴ AC^{2 }= AB^{2 }+ BC^{2}

∴ k^{2 }– 2k + 5 = 10 + k^{2 }– 4k + 5

∴ 2k = 10

∴ k = 5

Hence, if ∠B is a right angle in ∆ABC, then k = 5.

3. ∠C is a right angle.

∴ AB^{2 }= BC^{2 }+ AC^{2}

∴ 10 = k^{2 }– 4k + 5 + k^{2 }– 2k + 5

∴ 2k^{2 }– 6k = 0

∴ k^{2 }– 3k = 0

∴ k(k – 3) = 0

∴ k = 0 or k = 3

Hence, if ∠C is a right angle in ∆ABC, then k = 0 or k = 3

**Question 10:**

Show that the points P(3, -3), Q(-3, -3) and O(0, 0) are the vertices of an isosceles right angled triangle.

**Solution :**

### Exercise-8.2

**Question 1:**

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

**Solution :**

**Question 2:**

P(1, y) is a point on \(\overline{BC}\). (0, 2) and (3, 5) are the coordinates of A and B respectively. Find the ratio \(\frac{AP}{PB}\) and the value of y.

**Solution :**

**Question 3:**

Find the coordinates of points which divide the segment joining A(0, 0) and B(4, 8) in four congruent segments.

**Solution :**

**Question 4:**

If A(1, 2), B(2, 1), C(3, -4) are the vertices of \(\square \)^{m}ABCD, find the coordinates of D.

**Solution :**

**Question 5:**

(1, 1), (3, 2), (-1, 3) are the coordinates of mid-points of a triangle. Find the coordinates of the vertices of the triangle.

**Solution :**

**Question 6:**

A(X_{1}, Y_{1}), B(X_{2}, Y_{2}), (X_{3}, Y_{3}) are the vertices of ∆ABC and D, E, F are the mid-points of the sides \(\overline{BC}\), \(\overline{CA}\), \(\overline{AB}\) respectively. Prove that \(\square \) DEFB is a parallelogram.

**Solution :**

**Question 7:**

If A(-4, 3), B(10, 5), C(12, -9) are the vertices of a square, then find the coordinates of the fourth vertex.

**Solution :**

**Question 8:**

Find the coordinates of points which divide the segment joining A(-7, 5) and B(5, -1) into three congruent segments. (Such points are called the points of trisection of the segment).

**Solution :**

**Question 9:**

A(-4, 2), B(-2, 1), C(4, -2) are collinear points. Find the ratio in which B divides \(\overline{AC}\) from A.

**Solution :**

**Question 10:**

Show that A(2, 3), B(4, 5) and C(3, 2) can be the vertices of a rectangle. Find the coordinates of the fourth vertex.

**Solution :**

A(2, 3), B(4, 5) and C(3, 2) are the given points.

AB^{2 }= (2 – 4)^{2 }+ (3 – 5)^{2}

= 4 + 4

= 8

BC^{2 }= (4 – 3)^{2 }+ (5 – 2)^{2}

= 1 + 9

= 10

AC^{2 }= (2 – 3)^{2 }+ (3 – 2)^{2}

= 1 + 1

= 2

AB^{2 }+ AC^{2 }= 8 + 2

=10

∴ AB^{2 }+ AC^{2 }= BC^{2}

∴ In ∆ABC, ∠A is a right angle,

∴ ∆ABC is a right angled triangle. So, we can say that A, B and C can be the vertices of a rectangle whose fourth vertex is D(x, y) opposite to vertex A,

Now, the diagonals of a rectangle bisect each other

**Question 11:**

(2, 1), (-1, 2) and (1, 0) are the coordinates of three vertices of a parallelogram. Find the fourth vertex.

**Solution :**

Here (2, 1), (-1, 2) and (1, 0) are the three vertices of a parallelogram and the fourth vertex needs to be found.

The fourth vertex can be obtained in three different ways.

### Exercise-8.3

**Question 1:**

Find the area of ∆ABC whose vertices are A(4, 2), B(3, 9) and C(10, 10).

**Solution :**

**Question 2:**

Find a if the area of ∆ABC is 5. The coordinates A, B, C are (2, 3), (4, 5) and (a, 3).

**Solution :**

**Question 3:**

Find the area of \(\square \) ABCD where coordinates of A, B, C, D are (0, 0), (1, 2), (2, 5) and (1, 4)respectively.

**Solution :**

**Question 4:**

A(1, 5), B(3, -1), C(-5, 5) are the vertices of ∆ABC. Find the area of ∆DEF, if D, E, F are the mid-points of the sides of ∆ABC.

**Solution :**

**Question 5:**

(9, a), (6, 7), (2, 3) are the coordinates of the vertices of a triangle. If the area of the triangle is 10, find a.

**Solution :**

### Exercise-8

**Question 1:**

Can A(3, 4), B(0, -5), C(3, -1) be the vertices of a triangle ? If your answers is yes, find the area of the triangle. Hence find the length of the altitude on \(\overline{BC}\) .

**Solution :**

**Question 2:**

If A(5, 2), B(3, 4), C(x, y) are collinear and AB = BC, then find (x, y).

**Solution :**

**Question 3:**

Show that A(0, 1), B(2, 9) C\(\left( \frac{2}{3},\frac{11}{3} \right) \) are collinear. Find which point is between the remaining two. Also find the ratio in which \(\overline{AB}\) is divided by C from A.

**Solution :**

**Question 4:**

Find k if A(k, 2), B(3, 1), C(4, 2) are the vertices of a right angle triangle.

**Solution :**

A(k, 2), B(3, 1), C(4, 2) are the vertices of a right angles triangle.

AB^{2 }= (k – 3)^{2 }+ (2 – 1)^{2}

= k^{2 }– 6k + 9 + 1

= k^{2 }– 6k + 10

BC^{2 }= (3 – 4)^{2 }+ (1 – 2)^{2}

^{ }= 1 + 1

= 2

AC^{2 }= (k – 4)^{2 }+ (2 – 2)^{2}

^{ }= k^{2 }– 8k + 16

There are three possible cases:

Case 1:

m∠A = 90°

∴ BC^{2 }= AB^{2 }+ AC^{2}

∴ 2 = k^{2 }– 6k + 10 + k^{2 }– 8k + 16

∴ 2 = 2k^{2 }– 14k + 24

∴ 2k^{2 }– 14k + 24 = 0

∴ k^{2 }– 7k + 12 = 0

∴ k(k – 4) – 3(k – 4) = 0

∴ (k – 4)(k – 3) = 0

∴ k = 4 or k = 3. But for k = 4, the points A and C are equal so we cannot take k = 4. Thus, in this case we get k = 3.

Case 2:

m∠B = 90°

∴AC^{2 }= AB^{2 }+ BC^{2}

∴ k^{2 }– 8k + 16 = k^{2 }– 6k + 10 + 2

∴ 4 = 2k

∴ k = 2

Case 3:

m∠C = 90°

∴ AB^{2 }= BC^{2 }+ AC^{2}

∴ k^{2 }– 6k + 10 = 2 + k^{2 }– 8k + 16

∴ 2k = 8 ∴ k = 4

But for k = 4, A and C are equal. So k≠4

i.e., For the vertices A(k, 2), B(3, 1) and C(4, 2),

∠C = 90° is not possible.

Thus, for m∠A = 90°, we get, k = 3, for m∠B = 90°, we get, k = 2 and m∠C ≠ 90.

**Question 5:**

A(3, 0), B(0, 4) and C(3, 4) are the vertices of ∆ABC. The bisector of ∠C intersects \(\overline{AB}\) in D. Hint : The result\(\frac{AD}{DB}=\frac{CQ}{DA}\)

**Solution :**

**Question 6:**

A(3, -4), B(5, -2), C(-1, 8) are the vertices of ∆ABC. D, E, F are the mid-points of sides \(\overline{BC}\), \(\overline{CA}\) and \(\overline{AB}\) respectively. Find area of ∆ABC. Using coordinates of D, E, F, find area of ∆DEF. Hence show that the ABC = 4(DEF).

**Solution :**

**Question 7:**

Show that O(0, 0), A(a, 0), B(a, b), C(0, b) are the vertices of a rectangle. If P(x, y) is a point in the coordinate plane, then prove using distance formula that PO^{2} + PB^{2} = PA^{2} + PC^{2}.

**Solution :**

**Question 8:**

Find the coordinates of the points of intersection of the segment joining A(-3, -7) and B(3, 5) with Y-axis.

**Solution :**

**Question 9:**

A(1, 1), B(5, 4), C(3, 8), D(-1, 2) are the vertices of \(\square \) ABCD. If P, Q, R, S are the mid-points of \(\overline{AB}\), \(\overline{BC}\), \(\overline{CD}\), \(\overline{DA}\) respectively, show that \(\square \) PQRS is parallelogram.

**Solution :**

**Question 10:**

Select a proper option (a), (b), (c) or (d) from given options :

**Question 10(1):**

If A(1, 2) and B(3, -2) are given points, then …… is the mid-point of \(\overline{AB}\).

**Solution :**

**Question 10(2):**

One of the end-points of a circle having centre at origin is A(3, -2), then the other end-point of the diameter has the coordinates ……

**Solution :**

**Question 10(3):**

The distance of A(x, y) from origin is……….

**Solution :**

**Question 10(4):**

The foot of the perpendicular from P(-3, 2) to Y-axis is M. Coordinates of M are …..

**Solution :**

b. (0, 2)

It is clear from the following figure.

**Question 10(5):**

The coordinates of the foot of the perpendicular from P(5, -1) to X-axis are …..

**Solution :**

d. (5, 0)

This is clear from the following figure.

**Question 10(6):**

A(0, 0), B(3, 0), C(3, 4) are the vertices of a …… triangle.

**Solution :**

a. right angled

A(0, 0), B(3, 0) and C(3, 4)

AB^{2 }= (0 – 3)^{2 }+ (0 – 0)^{2}

= 9 + 0

= 9

BC^{2 }= (3 – 3)^{2 }+ (0 – 4)^{2}

= 0 + 16

=16

CA^{2 }= (3 – 0)^{2 }+ (4 – 0)^{2}

= 9 + 16

=25

Now,

AB^{2 }+ BC^{2}

= 9 + 16

= 25

∴ AB^{2 }+ BC^{2 }=CA^{2}

∴∆ABC is a right angled triangle and ∠B is a right angle.

**Question 10(7):**

(1, 0), (0, 1), (1, 1) are the coordinates of vertices of a triangle. The triangle is …… triangle.

**Solution :**

**Question 10(8):**

A(1, 2), B(2, 3), C(3, 4) are given points ……. of the following is true.

**Solution :**