# Class 10 RD Sharma Solutions – Chapter 14 Coordinate Geometry – Exercise 14.2 | Set 1

### Question 1. Find the distance between the following pair of points:

**(i) (-6, 7) and (-1,-5)**

**(ii) (a + b, b + c) and (a – b, c – b)**

**(iii) (a sin a, -b cos a) and (-a cos a, b sin a)**

**(iv) (a, 0) and (0, b)**

**Solution:**

(i)Given that P(-6, 7) and Q(-1, -5)So, x

_{1 }= -6, y_{1 }= 7x

_{2 }= -1, y_{2 }= -5Now we find the distance between PQ:

(ii)Given that P(a + b, b + c) and Q(a – b, c – b)So, x

_{1 }= a + b, y_{1 }= b + cx

_{2 }= a – b, y_{2 }= c – bNow we find the distance between PQ:

(iii)Given that P(a sin a,-b cos a) and Q(-a cos a, b sin a)So, x

_{1 }= a sin a, y_{1 }= -b cos ax

_{2 }= a cos a, y_{2 }= b sin aNow we find the distance between PQ:

(iv)Given that P(a, 0) and Q(0, b)So, x

_{1 }= a, y_{1 }= 0x

_{2 }= 0, y_{2 }= bNow we find the distance between PQ:

### Question 2. Find the value of a when the distance between the points (3, a) and (4, 1) is √10.

**Solution:**

Given that point P(3, a) and Q(4, 1) and the distance between them is √10

So, we have to find the value of a

So,

Squaring on both sides we get

(√10)

^{2 }= (√{2 + a^{2 }– 2a})^{2}10 = 2 + a

^{2 }– 2aa

^{2 }– 2a + 2 – 10 = 0a

^{2 }– 2a – 8 = 0On Splitting the middle term we get

a

^{2 }– 4a + 2a – 8 = 0a(a – 4) + 2(a – 4) = 0

(a – 4)(a + 2) = 0

a = 4, a = -2

### Question 3. If the points (2, 1) and (1, 2) are equidistant from the point(x, y) show that x + 3y = 0.

**Solution:**

Given that P(2, 1) and Q(1, -2) and R(x, y)

Also, PR = QR

x

^{2 }+ 5 – 4x + y^{2 }– 2y = x^{2 }+ 5 – 2x + y^{2 }+ 4yx

^{2 }+ 5 – 4x + y^{2 }– 2y = x^{2 }+ 5 – 2x + y^{2 }+ 4y-4x + 2x – 2y – 4y = 0

-2x – 6y = 0

-2(x + 3y) = 0

x + 3y = 0/-2

x + 3y = 0

Hence Proved

### Question 4. Find the values of x, y if the distance of the point(x, y) from(-3, 0) as well as from (3, 0) are 4.

**Solution:**

Given that P(x, y), Q(-3, 0) and R(3, 0).

Also, PQ = PR = 4

So,

On squaring on both sides, we get

(4)

^{2 }= (√x^{2 }+ 9 + 6x + y^{2})^{2}16 = x

^{2 }+ 9 + 6x + y^{2}x

^{2 }+ y^{2 }= 16 – 9 – 6xx

^{2 }+ y^{2 }= 7 – 6x ……..(1)On squaring on both sides, we get

(4)

^{2 }= (√x^{2 }+ 9 – 6x + y^{2})^{2}16 = x

^{2 }+ 9 – 6x + y^{2}x

^{2 }+ y^{2 }= 16 – 9 + 6xx

^{2 }+ y^{2 }= 7 + 6x ……..(2)From equation (1) and (2)

7 – 6x = 7 + 6x

7 – 7 = 6x + 6x

0 = 12x

x = 12

On substituting the value of x = 0 in eq(2)

x

^{2 }+ y^{2 }= 7 + 6x0 + y

^{2 }= 7 + 6 * 0y

^{2 }= 7y = ±√7

### Question 5. The length of a line segment is of 10 units and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the coordinate of the other end.

**Solution:**

Let the ordinate of other end by y, then The distance between (2, -3) and (10, y) is

= 10

Squaring on both sides we get

(8)

^{2 }+ (y + 3)^{2 }= 10064 + y

^{2 }+ 6y + 9 = 100y

^{2 }+ 6y + 73 – 100 = 0y

^{2 }+ 6y – 27 = 0y

^{2 }+ 9y – 3y – 27 = 0y(y + 9) – 3(y + 9) = 0

(y + 9)(y – 3) = 0

When y + 9 = 0, then y = -9

or when y – 3 = 0, then y = 3

So, the coordinates will be -9 or 3

### Question 6. Show that the points (-4, -1), (-2, -4), (4, 0), and (2, 3) are the vertices points of a rectangle.

**Solution:**

Let us considered ABCD is a rectangle whose vertices are

A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3)

Now

AB =

=

Similarly, CD =√13

AD = √52

and BC = √52

AC = √65 and BD = √65

Here, AB = CD and AD = BC

and diagonals AC = BD

So, ABCD is a rectangle

### Question 7. Show that the points A (1, -2), B (3, 6), C (5, 10), and D (3, 2) are the vertices of a parallelogram.

**Solution:**

Given points are A (1, -2), B (3, 6), C (5, 10) and D (3, 2)

Now AB =

Similarly BC =

So, from the above we conclude that AB = CD and AD = BC

Hence, ABCD is a parallelogram.

### Question 8. Prove that the points A (1, 7), B (4, 2), C (-1, -1), and D (-4, 4) are the vertices of a square.

**Solution:**

Given points are A (1, 7), B (4, 2), C (-1,-1), D (-4, 4)

If these are the vertices of a square, then its diagonals and sides are equal

AC =

So, AC = BD

Now AB=

So, AB = BC = CD = DA and diagonal AC = BD

Hence, the given figure ABCD is a square.

### Question 9.Prove that the points (3, 0), (6, 4), and (-1, 3) are the vertices of a right-angled isosceles triangle.

**Solution:**

Given points are A(3, 0), B(6, 4), and C(-1, 3)

Now we find the length of AB =

Similarly, BC=

From the above we conclude that AB = CA and BC is the longest side

Now we verify the Pythagoras theorem,

So, BC

^{2}= AB^{2 }+ CA^{2 }BC

^{2 }= (5)^{2}+ (5)^{2}BC

^{2 }= 25 + 2550

^{ }= 50So, AB

^{2 }+ CA^{2 }= BC^{2}Hence, the given triangle ABC is an isosceles right triangle.

### Question 10. Prove that (2, -2), (-2, 1), and (5, 2) are the vertices of a right-angled triangle. Find the area of the triangle and the length of the hypotenuse.

**Solution:**

Given points are A(2, -2), B(-2, 1) and C(5, 2)

Now we find the length of

We see that AB = CA and BC is the longest side.

Now we verify the Pythagoras theorem,

So, BC

^{2}= AB^{2 }+ CA^{2 }BC

^{2 }= (5)^{2}+ (5)^{2}BC

^{2 }= 25 + 2550

^{ }= 50So, AB

^{2 }+ CA^{2 }= BC^{2}So, the given triangle ABC is a right-angled triangle

Now we find the area of triangle ABC = 1/2 × Base × height

= 1/2 × 5 × 5

= 25/2 sq.units

And the length of the hypotenuse BC is √50.

### Question 11. Prove that the points (2a, 4a), (2a, 6a), and (2a + √3 a, 5a) are the vertices of an equilateral triangle.

**Solution:**

Given points are A(2a, 4a), B(2a, 6a) and C(2a + √3 a, 5a)

Now we find the length of

So, we conclude that the length of side AB = BC = CA = 2a

Hence, ∆ABC is equilateral triangle.

### Question 12. Prove that the points (2, 3), (-4, -6), and (1, 32) do not form a triangle.

Given points are A(2, 3), B(-4, -6), and C(1, 32)

Now we find the length of AB =

Similarly, for BC = √89 and CA = √2

As we know that the sum of two sides of triangle are always greater than the third side

So, BC + CA= √89 + √2 not greater than AB

Hence, the given points do not form a triangle.

### Question 13. The points A (2, 9), B (a, 5), and C (5, 5) are the vertices of a triangle ABC right-angled at B. Find the values of a and hence the area of ∆ABC.

**Solution:**

Given that, the points A (2, 9), B (a, 5) and C (5, 5) are the vertices of ∆ABC right-angled at B.

By Pythagoras theorem,

AC

^{2}= AB^{2}+ BC^{2}………(i)Now, by distance formula,

We find the length of AB =

=]

Now put the values of AB, BC and AC in equation(i), we get

25 = a

^{2 }– 4a + 20 + 25 + a^{2 }– 10a2a

^{2 }– 14a + 20 = 0a

^{2 }– 7a + 10 = 0a

^{2 }– 2a – 5a + 10 = 0a(a – 2) – 5(a – 2) = 0

(a – 5)(a – 5) = 0

a = 2, 5

Here, a ≠ 5, since at a = 5, the length of BC = 0. It is not possible because

the sides AB,BC and CA from a right angled triangle.

So, a = 2

Now, the coordinates are A (2, 9), B (2, 5) and C (5, 5)

Now we find the area of ∆ABC =

= 1/2[2(5 – 5) + 2(5 – 9) + 5(9 – 5)]

= 1/2(0 – 8 + 20)

= 1/2 × 12

= 6

Hence, the required area of ∆ABC is 6sq. units.

### Question 14. Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2, 3), and (-3, -2) is a rhombus.

**Solution:**

Given points are A(2, -1), B(3, 4), C(-2, 3), and D(-3, -2)

Now we find the length of sides AB, CD, DA, BD and diagonals AC, BD

Now we conclude that AB = BC = CD = DA = √26 and diagonal AC ≠ BD

Hence, ABCD is a rhombus

### Question 15. Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.

**Solution:**

Let us considered ABC is an isosceles triangle whose two vertices are A (2, 0) and B (2, 5)

So, the co-ordinates of third vertex C be (x, y)

And also given that AC = BC = 3

Now

On squaring both sides, we get

(x – 2)

^{2 }+ y^{2 }= 9x

^{2 }– 4x + 4 + y^{2}= 9x

^{2 }+ y^{2 }– 4x = 5 …….(i)Similarly,

On squaring both sides, we get

(x – 2)

^{2 }+ (y – 5)^{2 }= 9x

^{2 }– 4x + 4 + y^{2 }– 10y + 25 = 9x

^{2 }+ y^{2 }– 4x – 10y = -20 …….(ii)Now on subtracting eq(ii) from (i), we get

10y = 25

y = 25/10 = 5/2

On substituting the value of y in eq(i)

x

^{2 }– 4x + (5/2)^{2 }= 5x

^{2 }– 4x + 25/4 – 5 = 04x

^{2 }– 16x + 25 – 20 = 04x

^{2 }– 16x + 5 + 5 = 0Here a = 4, b = -16, c = 5

So, the co-ordinate of C will be (2 + √11/2, 5/2) or (2 – √11/2, 5/2)

### Question 16. Which point on x-axis is equidistant from (5, 9) and (-4, 6)?

**Solution:**

Let co-ordinates of two points are A (5, 9), B (-4, 6)

The required point is on x-axis

Its ordinates or y-co-ordinates will be 0

Let the co-ordinates of the point C be (x, 0)

AC = CB

Now

Squaring both sides, we get

(x – 5)

^{2 }+ 81 = (x + 4)^{2 }+ 36x

^{2 }– 10x + 25 + 81 = x^{2 }+ 8x + 16 + 36 – 10x – 8x-18 = 52 – 106

-18x = -54

x = -54/18

x = 3

Hence, the required point is (3, 0)

### Question 17. Prove that the point (-2, 5), (0, 1) and (2, -3) are collinear.

**Solution:**

Given points are A(-2, 5), B(0, 1) and C(2, -3)

Now we find the length of AB, BC, and CA

Now AB + BC = 2√5 + 2√5

And CA = 4√5

AB + BC = CA

Hence, A, B and C are collinear

### Question 18. The coordinates of the point P are (-3, 2). Find the coordinates of the point Q which lies on the line joining P and origin such that OP = OQ.

**Solution:**

Given that the co-ordinates of P are (-3, 2) and origin O are (0, 0)

Let us assume that the co-ordinates of Q be (x, y)

Here, O is the mid-point of line PQ

so by using mid point formula we get,

(x – 3)/2 = 0 and (y + 3)/2 = 0

x = 3, y = -2

Hence, the coordinates of the point Q are (3, -2)

### Question 19. Which point on y-axis is equidistant from (2, 3) and (-4, 1)?

**Solution:**

The required point lies on y-axis

Its abscissa will be zero

So, let us assume that the point be C (0, y) and A (2, 3), B (-4, 1)

Now, we find the length of AC and BC

Here, we conclude that AC = BC

So,

On squaring both sides, we get

4 + (y – 3)

^{2 }= 16 + (y – 1)^{2}4 + y

^{2 }+ 9 – 6y = 16 + y^{2 }+ 1 – 2y-6y + 2y = 17 – 13

-4y = 4 = y = 4/-4 = 1

Hence, the coordinates of the required point is (0,-1)

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