Question

63. Given points A(1,5), B(-3, 7) and C(15,9).
(1) Find the equation of a line passing through the mid-point of AC and the point B.
(in) Find the equation of the line through C and parallel to AB.
(iii) The lines obtained in parts (i) and (ii) above, intersect each other at a point P. Find
the co-ordinates of the point P.
(iv) Assign, giving reason, a special names of the figure PABC.​

Answer 1

Step-by-step explanation:

63. Given points A(1,5), B(-3, 7) and C(15,9).

(1) Find the equation of a line passing through the mid-point of AC and the point B.

Mid point of AC

$x = \frac{1 + 15}{2} = 8 \\ \\ y = \frac{9 + 5}{2} = 7$

Equation of a line passing through two given points

$y - y1 = \frac{y2 - y1}{x2 - x1} (x - x1) \\ \\ y - 7 = \frac{7 - 7}{ 8 + 3} (x + 3) \\ \\ y - 7 = 0$

2) Find the equation of the line through C and parallel to AB.

Slope of AB

$m = \frac{7 - 5}{ - 3 - 1} \\ \\ m = \frac{2}{ - 4} \\ \\ m = \frac{ - 1}{2}$

Equation of line

$y - 9 = \frac{ - 1}{2} (x - 15) \\ \\ 2y - 18 = - x + 15 \\ \\ x + 2y = 33..eq2$

(iii) The lines obtained in parts (i) and (ii) above, intersect each other at a point P. Find

the co-ordinates of the point P.

$x + 14 = 33 \\ \\ x = 33 - 14 \\ \\ x = 19$

coordinate of P(19,7)

Hope it helps you.

Answer 2

Given : points A(1, 5), B(-3, 7) and C(15, 9).

To Find :  

the equation of a line passing through the mid-point of AC and the point B.

equation of the line through C and parallel to AB

Find intersection point P

Solution:

A(1, 5)  and C(15, 9).

Mid point of A and C  = ( 1 + 15)/2 , ) (5 + 9)/2

= 8 , 7

B(-3, 7)

(8 , 7)

slope  =   0

y - 7   = 0 (x - 8)

=> y = 7

y = 7 is the equation of a line passing through the mid-point of AC and the point B.

equation of the line through C and parallel to AB

A(1, 5) B(-3, 7)  

Slope =  (7 - 5)/(-3 - 1)  = 2/(-4) = -1/2

C(15, 9).

y  - 9 = (-1/2)(x - 15)

=> 2y - 18 = -x + 15

=> x + 2y = 33

y = 7

x + 2y = 33

=> x = 19

P= ( 19 , 7)

PABC is parallelogram

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