Question

The line $x-4y=6$ is the perpendicular bisector of the line segment $AB$ If $B=(1,3)$, find the co-ordinates of point $A$.

Answer 1

length of rectangle = 3k - 1

breadth of rectangle = 2k -1

Given area = 144

(3k-1)(2k-1) = 144

$\Rightarrow 6k^2-5k+1=144$

Given perpendicular b sector of line

segment AB x - 4y = 6

Given B = (1,3)

slope of AB = $\frac{1}{4}$

M slope $\perp$ AB = -4

line equation $\Rightarrow$$y-3=-4(n-1)$

y-3 = -4n+4

4x + y -7 = 0

16x + 4y -28 = 0

x -4y -6 = 0

$17x-34=0$ $\Rightarrow$ x = 2

4y = -4$\Rightarrow$ y = -1

mid point of AB = (2,-1)

(2,-1)$(\frac{x^a+1}{2},\frac{y^A+3}{2})$ $\Rightarrow x_A=3y_A=-5$

A$\Rightarrow$(3,-5)