Question

(6,1) ,(8,2) ,(9,x) ,(p,3) are the vertices of a parallelogram. Find the values of p and x.

Answer 1

Let A = (6 , 1)
B = (8, 2)
C = (9, x)
D = (P , 3)

we know,
properties of parallelogram ,
mid point of diagonal AC = midpoint of diagonal BD

use section formula,for finding mid point
$(x,y)=[\frac{x_1+y_1}{2},\frac{x_2+y_2}{2}]$

Mid point of AC = {(9 + 6)/2 , (1 + x)/2 }
mid point of BD = {(8+P)/2 , (3 + 2)/2}
so, for x- co-ordinate
(9 + 6)/2 = (8+P)/2
15 = 8 + P
P = 7

similarly, for y - co-ordinate
(1 + x)/2 = (3 + 2)/2
1 + x = 5
x = 4

Answer 2

In parallelogram diagonal bisects each other.

Therefore

midpoint of AC = midpoint of BD

$( \frac{6 + 9}{2}, \frac{1 + x}{2} ) = ( \frac{8 + p}{2}, \frac{2 + 3}{2} )$

Therefore,

$\frac{6 + 9}{2} = \frac{8 + p}{2}$

$15 = 8 + p$

$p = 15 - 8 = 7$
Now, In same way

$\frac{1 + x}{2} = \frac{2 + 3}{2}$

$1 + x = 5$

$x = 5 - 1 = 4$
Answer:- x = 4 and p = 7