GSEB Solutions for Class 10 mathematics – Coordinate Geometry

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

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Exercise-8.1

Question 1:

Find the distance between the following pairs of points :

1. (2, -3), (7, 9)
2. (-3, 4), (0, 0)
3. (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
∴ PA² = 169
∴ (x – 11)² + (0 – 12)² = 169
∴ x² – 22x + 121 + 144 = 169
∴ x² – 22x + 265 = 169
∴ x² – 22x + 96 = 0
∴ x² – 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 + a² + 36 = 100
∴ a² – 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,

1. Find k, if ∠A is a right angle
2. Find k, if ∠B is a right angle
3. 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.
AB² = (1 – 2)² + (7 – 4)² = 1 + 9 = 10
BC² = (2 – k)² + (4 – 5)²
= 4 – 4k + k² + 1
= k² – 4k + 5
AC² = (1 – k)² + (7 – 5)²
= 1 – 2k + k² + 4
= k² – 2k + 5
1. ∠A is a right angle.
∴ BC² = AB² +AC²
∴ k² – 4k + 5 = 10 + k² – 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.
∴ AC² = AB² + BC²
∴ k² – 2k + 5 = 10 + k² – 4k + 5
∴ 2k = 10
∴ k = 5
Hence, if ∠B is a right angle in ∆ABC, then k = 5.

3. ∠C is a right angle.
∴ AB² = BC² + AC²
∴ 10 = k² – 4k + 5 + k² – 2k + 5
∴ 2k² – 6k = 0
∴ k² – 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 \(\square \)^mABCD, 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(X₁, Y₁), B(X₂, Y₂), (X₃, Y₃) 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.
AB² = (2 – 4)² + (3 – 5)²
= 4 + 4
= 8
BC² = (4 – 3)² + (5 – 2)²
= 1 + 9
= 10
AC² = (2 – 3)² + (3 – 2)²
= 1 + 1
= 2
AB² + AC² = 8 + 2
=10
∴  AB² + AC² = BC²
∴ 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.
AB² = (k – 3)² + (2 – 1)²
= k² – 6k + 9 + 1
= k² – 6k + 10
BC² = (3 – 4)² + (1 – 2)²
 = 1 + 1
= 2
AC² = (k – 4)² + (2 – 2)²
 = k² – 8k + 16
There are three possible cases:
Case 1:
m∠A = 90°
∴ BC² = AB² + AC²
∴ 2 = k² – 6k + 10 + k² – 8k + 16
∴ 2 = 2k² – 14k + 24
∴ 2k² – 14k + 24 = 0
∴ k² – 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°
∴AC² = AB² + BC²
∴ k² – 8k + 16 = k² – 6k + 10 + 2
∴ 4 = 2k
∴ k = 2
Case 3:
m∠C = 90°
∴ AB² = BC² + AC²
∴ k² – 6k + 10 = 2 + k² – 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 PO² + PB² = PA² + PC².

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)
AB² = (0 – 3)² + (0 – 0)²
= 9 + 0
= 9
BC² = (3 – 3)² + (0 – 4)²
= 0 + 16
=16
CA² = (3 – 0)² + (4 – 0)²
= 9 + 16
=25
Now,
AB² + BC²
= 9 + 16
= 25
∴ AB² + BC² =CA²
∴∆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 :