# RS Aggarwal Class 10 Solutions Chapter 16 - Coordinate Geometry Ex 16F (16.6)

## RS Aggarwal Class 10 Chapter 16 - Coordinate Geometry Ex 16F (16.6) Solutions Free PDF

Question 1:

Points A(-1, y), B(5, 7) lie on a circle with centre O(2, -3y). Find the values of y.

Solution:

The given points are A(-1, y), B(5, 7) and O(2, -3y).

Here, AO and BO are the radii of the circle. So

AO = BO

Hence, AO2 = BO2

$\Rightarrow$ = (2 + 1)2 + (-3y – y)2 = (2 – 5)2 + (-3y – 7)2

$\Rightarrow$ 9 + (4y)2 = (-3)2 + (3y + 7)2

$\Rightarrow$ 9 + 16y2 = 9 + 9y2 + 49 + 42y

$\Rightarrow$ 7y2 – 42y – 49 = 0

$\Rightarrow$ y2 – 6y – 7 = 0

$\Rightarrow$ y2 – 7y + y – 7 = 0

$\Rightarrow$ y(y – 7) + 1(y – 7) = 0

$\Rightarrow$ (y – 7) (y + 1) = 0

$\Rightarrow$ y = -1 or y = 7

Hence, y = 7 or y = -1

Question 2:

If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find P.

Solution:

The given points are A(0, 2), B(3, p) and C(p, 5)

AB = AC $\Rightarrow$ AB2 = AC2

$\Rightarrow$ (3 – 0)2 + (p – 2)2 = (p – 0)2 + (5 – 2)2

$\Rightarrow$ 9 + p2 – 4p + 4 = p2 + 9

$\Rightarrow$ 4p = 4 $\Rightarrow$ p = 1

Hence, p = 1.

Question 3:

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Find the length of one of its diagonal.

The given vertices are B(4, 0) , C(4, 3) and D(0, 3). Here, BD is one of the diagonals. So,

$BD = \sqrt{(4 – 0)^{2} + (0 – 3)^{2}}$

$\Rightarrow \sqrt{4^{2} + (-3)^{2}}$

$\Rightarrow \sqrt{16 + 9}$

$\Rightarrow = \sqrt{25} = 5$

Hence, the length of the diagonal is 5 units

Question 4:

If the point P(k-1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.

The given points are P(k-1, 2), A(3, k) and B(k, 5)

AP = BP $\Rightarrow$ AP2 = BP2

$\Rightarrow$ (k -1 – 3)2 + (2 – k)2 = (k – 1 – k)2 + (2 – 5)2

$\Rightarrow$ (k – 4)2 + (2 – k)2 = (-1)2 + (-3)2

$\Rightarrow$ k2 – 8y + 16 + 4 + k2 – 4k = 1+ 9

$\Rightarrow$ k2 – 6y + 5 = 0

$\Rightarrow$ (k – 1) (k – 5) = 0

$\Rightarrow$ k = 1 or k = 5

Hence, k = 1 or k = 5.

Question 5:

Find the ratio in which the point P(x, 2) divides the join of A(12, 5) and B(4, -3).

Solution:

Let k be the ratio in which the point P(x, 2) divides the line joining the points A(x1 = 12, y1 = 5) and B(x2 = 4, y2 = -3) then,

$x = \frac{k \times 4 + 12}{k + 1}$ and $2 = \frac{k \times (-3) + 5}{k + 1}$

Now,

$2 = \frac{k \times (-3) + 5 }{k + 1} \Rightarrow 2k + 2 = -3k + 5 \Rightarrow k = \frac{3}{5}$

Hence, the required ration is 3:5.

Question 6:

Prove that the diagonals of a rectangle ABCD with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6) are equal and bisect each other.

The vertices of the rectangle ABCD are A(2, -1) , B(5, -1), C(5, 6) and D(2, 6). Now,

Coordinates of midpoint of AC = $\frac{2 + 5}{2}, \frac{-1 + 6}{2} = \frac{7}{2}, \frac{5}{2}$

Coordinates of midpoint of BD = $\frac{5 + 2}{2}, \frac{-1 + 6}{2} = \frac{7}{2}, \frac{5}{2}$

Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.

Question 7:

Find the lengths of the medians AD and BE of triangle ABC whose vertices are A(7, -3), B(5, 3) and C(3, -1).

The given vertices are A(7, -3), B(5, 3) and C(3, -1)

Since D and E are the midpoints of BC and AC respectively, therefore

Coordinates of D = $\frac{5 + 3}{2} , \frac{3 – 1}{2} = (4, 1)$

Coordinates of E = $\frac{7 + 3}{2} , \frac{-3 -1}{2} = (5, -2)$

Now,

$AD = \sqrt{(7 – 4)^{2} + (-3 -1)^{2}} = \sqrt{9 + 16} = 5$

$BE = \sqrt{(5 – 5)^{2} + (3 + 2)^{2}} = \sqrt{0 + 25} = 5$

Hence, AD = BE = 5 units

Question 8:

If the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2:3 then find the value of K.

Here, the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2:3. So,

$k = \frac{2 \times 5 + 3 \times 2}{2 + 3}$

= $\frac{10 + 6}{5}$

$= \frac{16}{5}$

Hence, $k = \frac{16}{5}$.

Question 9:

Find the point on x-axis which is equidistant from points A(-1, 0) and B(5, 0).

Let p(x, 0) be the point on x-axis. Then,

AP = BP $\Rightarrow$ AP2 = BP2

$\Rightarrow$ (x + 1)2 Â + (0 – 0)2 = (x – 5)2 + (0 – 0)2

$\Rightarrow$ x2 + 2x + 1 = x2 – 10x + 25

$\Rightarrow$ 12x = 24 $\Rightarrow$ x = 2

Hence, x = 2.

Question 10:

Find the distance between the points $A(\frac{-8}{5}, 2) and B(\frac{2}{5}, 2)$

The given points are $A(\frac{-8}{5}, 2) and B(\frac{2}{5}, 2)$

Then, $(x_{1} = \frac{-8}{5}, y_{1} = 2) and (x_{2} = \frac{2}{5}, y_{2} = 2)$

Therefore,

$AB = \sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}}$

= $\sqrt{{\frac{2}{5} – \frac{-8}{5}}^{2} + (2 – 2)^{2}}$

= $\sqrt{2^{2} + 0^{2}}$

= $\sqrt{4 + 0}$

= $\sqrt{4} = 2 \, units$

Question 11:

Find the value of a, so that the point (3, a) lies on the line represented by 2x-3y = 5.

The point (3, a) lies on the line 2x -3y = 5

If point (3, a) lies on the line 2x -3y = 5, then

2x â€“ 3y = 5

$\Rightarrow$ (2*3) â€“ (3*a) = 5

$\Rightarrow$ 6 â€“ 3a = 5

$\Rightarrow a=\frac{1}{3}$

Hence, the value of a is $a=\frac{1}{3}$

Question 12:

If the points A(4,3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.

The given points A (4, 3) and B (x, 5) lie on the circle with center O (2, 3)

Then, OA = OB

$\Rightarrow \sqrt{(x-2)^{2}+(5-3)^{2}} = \sqrt{(4-2)^{2}+(3-3)^{2}}$

$\Rightarrow (x-2)^{2} + 2^{2} = 2^{2} + 0^{2}$

$\Rightarrow (x-2)^{2} = (2^{2} – 2^{2})$

$\Rightarrow (x-2)^{2} = 0$

$\Rightarrow (x-2) = 0$

$\Rightarrow x = 2$

Hence, the value of x = 2

Question 13:

If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.

Let the point P (x, y) be equidistant from the points A (7, 1) and B (3, 5)

Then, PA = PB

$\Rightarrow PA^{2}=PB^{2}$

$\Rightarrow \sqrt{(x-7)^{2}+(y-1)^{2}} = \sqrt{(x-3)^{2}+(y-5)^{2}}$

$\Rightarrow x^{2} + y^{2} – 14x – 2y + 50 = x^{2} +y^{2} -6x -10y +34$

$\Rightarrow 8x â€“ 8y = 16$

$\Rightarrow x â€“ y = 2$

Question 14:

If the centroid of triangle ABC having vertices A(a, b), B(b, c) and C(c, a) is the origin, then find the value of (a + b + c).

The given points are A (a, b), B (b, c) and C (c, a)

Here,

(x1 = a, y1 = b), (x2 = b, y2 = c) and (x3 = c, y3 = a)

Let the centroid be (x, y)

Then,

$x=\frac{1}{3}(x_{1}+x_{2}+x_{3})$

$x=\frac{1}{3}(a +b + c)$

= $\frac{a+b+c}{3}$

$y =\frac{1}{3}(y_{1}+ y_{2}+ y_{3})$

$y =\frac{1}{3}(b + c + a)$

= $\frac{a+b+c}{3}$

But it is given that the centroid of the triangle is the origin

Then, we have:

$\frac{a+b+c}{3}=0$

$\Rightarrow a+b+c=0$

Question 15:

Find the centroid of triangle ABC whose vertices are A(2,2), B (-4, -4) and C (5, -8)

The given points are A (2, 2), B (-4, -4) and C (5, -8)

Here,

(x1 = 2, y1 = 2), (x2 = -4, y2 = -4) and (x3 = 5, y3 = -8)

Let G (x, y) be the centroid of $\Delta ABC$.Then,

$x=\frac{1}{3}(x_{1}+x_{2}+x_{3})$

=$\frac{1}{3}(2 – 4 + 5)$

= 1

$y=\frac{1}{3}(y_{1}+ y_{2}+ y_{3})$

= $\frac{1}{3}(2 – 4 – 8)$

= $\frac{-10}{3}$

Hence, the centroid of $\Delta ABC$ is G$\left (1,\frac{-10}{3}\right )$

Question 16:

In what ratio does the point C(4,5) divide the join of A(2, 3) and B(7, 8) ?

Let the required ratio be k: 1

Then, by section formula, the co-ordinates of C are

C$\left (\frac{7k+2}{k+1},\frac{8k+3}{k+1} \right )$

Therefore,

$\frac{7k+2}{k+1}=4\;and\;\frac{8k+3}{k+1}=5$$\left [Because, C(4,5)\;is\;given\right ]$

$\Rightarrow$ 7k + 2 = 4k + 4 and 8k + 3 = 5k +5

$\Rightarrow$ 3k = 2 and 3k = 2

$\Rightarrow k=\frac{2}{3}$ in each case

So, the required ratio is $\frac{2}{3}:1$, which is same as 2:3

Question 17:

If the points A(2, 3) , B(4, k) and C(6, -3) are collinear, find the value of k.

The given points are A(2, 3) , B(4, k) and C(6, -3)

Here

(x1 = 2 , y1 = 3), (x2 = 4 , y2 = k) and (x3 = 6 , y3 = -3)

It is given that the points A, B and C are collinear. Then

x1(y2 â€“ y3) + x2(y3 â€“ y1) + x3(y1 â€“ y2) = 0

$\Rightarrow$ 2(k + 3) + 4(-3 â€“ 3) +6(3 â€“ k) = 0

$\Rightarrow$ 2k + 6 -24 +18 -6k = 0

$\Rightarrow$ -4k = 0

$\Rightarrow$ k = 0′

#### Practise This Question

Far point of a normal eye is situated at