CBSE 10th Maths -  Coordinate Geometry - Exercise 7.4

Solution:

Consider line 2x + y – 4 = 0 divides line AB joined by the two points A(2, -2) and B(3, 7) in k:1 ratio.

Coordinates of point of division can be given as follows:

x = $\dfrac{(2 + 3k)}{(k + 1)}$ and y = $\dfrac{(-2 + 7k)}{(k + 1)}$

Substituting the values of x and y given equation, i.e. 2x + y – 4 = 0, we have

2{ $\dfrac{(2 + 3k)}{(k + 1)}$} + { $\dfrac{(-2 + 7k)}{(k + 1)}$} – 4 = 0

$\dfrac{(4 + 6k)}{(k + 1)}$ + $\dfrac{ (-2 + 7k)}{(k + 1)}$= 4

4 + 6k – 2 + 7k = 4(k+1)

-2 + 9k = 0

Or k = $\dfrac{2}{9}$

Hence, the ratio is 2:9.

Solution:

If given points are collinear, then the area of the triangle formed by them must be zero.

Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,

Area of a triangle = $\dfrac{1}{2} × [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]$ = 0

[x(2 – 0) + 1 (0 – y) + 7( y – 2)] = 0

2x – y + 7y – 14 = 0

2x + 6y – 14 = 0

x + 3y – 7 = 0.

Which is the required result.

Solution:

Let A = (6, -6), B = (3, -7), and C = (3, 3) are the points on a circle.

If O is the centre, then OA = OB = OC (radii are equal)

If O = (x, y), then

OA = $\sqrt{[(x – 6)^{2} + (y + 6)^{2}]}$

OB = $\sqrt{[(x – 3)^{2}+ (y + 7)^{2}]}$

OC = $\sqrt{[(x – 3)^{2} + (y – 3)^{2}]}$

Choose: OA = OB, we have

After simplifying above, we get -6x = 2y – 14 ….(1)

Similarly, OB = OC

$(x – 3)^{2} + (y + 7)^{2}$ = $(x – 3)^{2} + (y – 3)^{2}$

$(y + 7)^{2}$= $(y – 3)^{2}$

$y^{2} + 14y + 49 $= $y^{2}– 6y + 9$

20y =-40

or y = -2

Substituting the value of y in equation (1), we get

-6x = 2y – 14

-6x = -4 – 14 = -18

x = 3

Hence, the centre of the circle is located at point (3,-2).

Solution:

Let ABCD is a square, where A(-1,2) and B(3,2). And Point O is the point of intersection of AC and BD.

To Find: Coordinate of points B and D.

Step 1: Find the distance between A and C and the coordinates of point O.

We know that the diagonals of a square are equal and bisect each other.

AC = $\sqrt{[(3 + 1)^{2} + (2 – 2)^{2}]}$ = 4

Coordinates of O can be calculated as follows:

x = $\dfrac{(3 – 1)}{2}$ = 1 and y = $\dfrac{(2 + 2)}{2}$ = 2

So, O(1,2)

Step 2: Find the side of the square using the Pythagoras theorem

Let a be the side of the square and AC = 4

From the right triangle, ACD,

a = 2$\sqrt{2}$

Hence, each side of the square = 2$\sqrt{2}$

Step 3: Find the coordinates of point D

Equate the length measure of AD and CD

Say, if the coordinates of D are $(x_{1}, y_{1})$

AD = $\sqrt{[(x_{1} + 1)^{2} + (y_{1} – 2)^{2}]}$

Squaring both sides,

$AD^{2}$ = $(x_{1} + 1)^{2}+ (y_{1} – 2)^{2}$

Similarly, $CD^{2}$ = $(x_{1} – 3)^{2} + (y_{1} – 2)^{2}$

Since all sides of a square are equal, which means AD = CD

$(x_{1} + 1)^{2} + (y_{1} – 2)^{2} = (x_{1} – 3)^{2} + (y_{1} – 2)^{2}$

$(x_{1})^{2} + 1 + 2x_{1}$ = $(x_{1})^{2} + 9 – 6x_{1}$

$8x_{1}$ = 8

$x_{1}$ = 1

The value of $y_{1}$ can be calculated as follows by using the value of x.

From step 2: each side of the square = 2$\sqrt{2}$

$CD^{2}$ = $(x_{1} – 3)^{2}+ (y_{1} – 2)^{2}$

8 = $(1 – 3)^{2} + (y_{1} – 2)^{2}$

8 =$ 4 + (y_{1} – 2)^{2}$

$y_{1} – 2$ = 2

$y_{1}$ = 4

Hence, D = (1, 4)

Step 4: Find the coordinates of point B

From line segment, BOD

Coordinates of B can be calculated using coordinates of O, as follows:

Earlier, we had calculated O = (1, 2)

Say B = $(x_{2}, y_{2})$

For BD:

1 = $\dfrac{(x_{2} + 1)}{2}$

$x_{2}$ = 1

And 2 = $\dfrac{(y_{2} + 4)}{2}$

=> $y_{2}$ = 0

Therefore, the coordinates of required points are B = (1,0) and D = (1,4)

Solution:

(i) Taking A as the origin, the coordinates of the vertices P, Q and R are,

From figure: P = (4, 6), Q = (3, 2), R (6, 5)

Here, AD is the x-axis and AB is the y-axis.

(ii) Taking C as the origin,

The coordinates of vertices P, Q and R are ( 12, 2), (13, 6) and (10, 3), respectively.

Here, CB is the x-axis and CD is the y-axis.

Find the area of triangles:

Area of triangle PQR in case of origin A:

Using formula: Area of a triangle = $\dfrac{1}{2} × [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]$

= $\dfrac{1}{2} [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]$

= $\dfrac{1}{2}$ (- 12 – 3 + 24 )

= $\dfrac{9}{2}$ sq unit

Area of triangle PQR in case of origin C:

Area of a triangle = $\dfrac{1}{2}$ × $[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]$

= $\dfrac{1}{2}$ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]

= $\dfrac{1}{2}$( 36 + 13 – 40)

= $\dfrac{9}{2}$ sq unit

This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C

The area is the same in both cases because the triangle remains the same no matter which point is considered as the origin.

Solution:

Given: The vertices of a $\triangle$ ABC are A (4, 6), B (1, 5) and C (7, 2)

$\dfrac{AD}{AB}$ = $\dfrac{AE}{AC}$ = $\dfrac{1}{4}$

$\dfrac{AD}{(AD + BD)}$ = $\dfrac{AE}{(AE + EC)}$ = $\dfrac{1}{4}$

Point D and Point E divide AB and AC, respectively, in ratio 1:3.

Coordinates of D can be calculated as follows:

x = $\dfrac{(m_{1}x_{2} + m_{2}x_{1})}{(m_{1} + m_{2})}$ and y = $\dfrac{(m_{1}y_{2}+ m_{2}y_{1})}{(m_{1} + m_{2})}$

Here, $m_{1}$ = 1 and $m_{2}$ = 3

Consider line segment AB which is divided by point D at the ratio 1:3.

x = $\dfrac{[3(4) + 1(1)]}{4}$ = $\dfrac{13}{4}$

y =$\dfrac{[3(6) + 1(5)]}{4}$ = $\dfrac{23}{4}$

Similarly, the coordinates of E can be calculated as follows:

x =$\dfrac{ [1(7) + 3(4)]}{4}$ = $\dfrac{19}{4}$

y =$\dfrac{ [1(2) + 3(6)]}{4}$ = $\dfrac{20}{4}$ = 5

Find the area of triangle:

Using formula: Area of a triangle = $\dfrac{1}{2}$ × $[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]$

The area of triangle $\triangle$ ABC can be calculated as follows:

= $\dfrac{1}{2}$ [4(5 – 2) + 1( 2 – 6) + 7( 6 – 5)]

= $\dfrac{1}{2}$ (12 – 4 + 7) = 15/2 sq unit

The area of $\triangle$ ADE can be calculated as follows:

= $\dfrac{1}{2} [4(\dfrac{23}{4} – 5) + (\dfrac{13}{4} (5 – 6) + (\dfrac{19}{4} (6 – \dfrac{23}{4})]$

= $\dfrac{1}{2} (3 – \dfrac{13}{4} + \dfrac{19}{16} )$

= $\dfrac{1}{2} ( \dfrac{15}{16} ) = \dfrac{15}{32}$ sq unit

Hence, the ratio of the area of triangle ADE to the area of triangle ABC = 1:16.

Solution:

(i) Coordinates of D can be calculated as follows:

Coordinates of D = ( $\dfrac{(6+1)}{2}, \dfrac{(5+4)}{2}$ ) = ($\dfrac{7}{2}, \dfrac{9}{2}$)

So, D is ($\dfrac{7}{2}, \dfrac{9}{2}$)

(ii) Coordinates of P can be calculated as follows:

Coordinates of P = ( $\dfrac{[2(\dfrac{7}{2}) + 1(4)]} {(2 + 1)}, \dfrac{[2(\dfrac{9}{2}) + 1(2)]}{(2 + 1)}$ ) = ($\dfrac{11}{3}, \dfrac{11}{3}$)

So, P is ($\dfrac{11}{3}, \dfrac{11}{3}$)

(iii) Coordinates of E can be calculated as follows:

Coordinates of E = ( $\dfrac{(4+1)}{2}, \dfrac{(2+4)}{2} $) =$ (\dfrac{5}{2}, \dfrac{6}{2})$ = $(\dfrac{5}{2} , 3)$

So, E is $(\dfrac{5}{2} , 3)$

Points Q and P would be coincident because the medians of a triangle intersect each other at a common point called the centroid. Coordinate of Q can be given as follows:

Coordinates of Q =( $\dfrac{[2(\dfrac{5}{2}) + 1(6)]}{(2 + 1)}, \dfrac{[2(3) + 1(5)]}{(2 + 1)}$) = ($\dfrac{11}{3}, \dfrac{11}{3}$)

F is the midpoint of the side AB

Coordinates of F = ( $\dfrac{(4+6)}{2}, \dfrac{(2+5)}{2}$ ) = (5, $\dfrac{7}{2}$)

Point R divides the side CF in ratio 2:1

Coordinates of R = ( $\dfrac{[2(5) + 1(1)]}{(2 + 1)}$, $\dfrac{[2(\dfrac{7}{2}) + 1(4)]}{(2 + 1)}$ ) =($\dfrac{11}{3}, \dfrac{11}{3}$)

(iv) Coordinates of P, Q and R are the same, which shows that medians intersect each other at a common point, i.e. centroid of the triangle.

(v) If A $(x_{1}, y_{1})$, B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ are the vertices of triangle ABC, the coordinates of the centroid can be given as follows:

x = $\dfrac{(x_{1} + x_{2} + x_{3})}{3}$ and y = $\dfrac{(y_{1} + y_{2} + y_{3})}{3}$

Solution:

P id the midpoint of side AB,

Coordinate of P = ( $\dfrac{(-1 – 1)}{2}, \dfrac{(-1 + 4)}{2}$ ) = (-1, $\dfrac{3}{2}$)

Similarly, Q, R and S are (As Q is the midpoint of BC, R is the midpoint of CD and S is the midpoint of AD)

Coordinate of Q = (2, 4)

Coordinate of R = (5, $\dfrac{3}{2}$)

Coordinate of S = (2, -1)

Now,

Length of PQ = $\sqrt{[(-1 – 2)^{2} + (\dfrac{3}{2} – 4)^{2}]}$ = $\sqrt{(\dfrac{61}{4})}$ = $\sqrt{\dfrac{61}{2}}$

Length of SP = $\sqrt{[(2 + 1)2 + (-1 – \dfrac{3}{2})^{2}]}$ = $\sqrt{(\dfrac{61}{4})}$= $\sqrt{\dfrac{61}{2}}$

Length of QR = $\sqrt{[(2 – 5)^{2} + (4 – \dfrac{3}{2})^{2}]}$ = $\sqrt{(\dfrac{61}{4})}$ = $\sqrt{\dfrac{61}{2}}$

Length of RS = $\sqrt{[(5 – 2)2 + (\dfrac{3}{2} + 1)^{2}]}$ = $\sqrt{(\dfrac{61}{4})}$ = $\sqrt{\dfrac{61}{2}}$

Length of PR (diagonal) =$\sqrt{[(-1 – 5)^{2} + (\dfrac{3}{2} – \dfrac{3}{2})^{2}]}$ = 6

Length of QS (diagonal) = $\sqrt{[(2 – 2)^{2} + (4 + 1)^{2}]}$ = 5

The above values show that PQ = SP = QR = RS = $\sqrt{\dfrac{61}{2}}$, i.e. all sides are equal.

But PR ≠ QS, i.e. diagonals are not of equal measure.

Hence, the given figure is a rhombus.