Question

One of the diagonals of a square is the portion of the line $\frac{x}{2}+\frac{y}{3}=2$ intercepted between the axes. Then the extremities of the other diagonal are

• A. $(5,5),(-1,1)$
• B. $(0,0),(4,6)$
• C. $(0,0),(-1,1)$
• D. $(5,5),(4,6)$

Answer 1

The correct option is A $(5,5),(-1,1)$

Let the quadrilateral be ABCD,

Given diagonal is AC.

Extremities of the given diagonal are (4, 0) and (0, 6).
Using the concept line perpendicular to $ax+by+c=0$ is $bx-ay+c^{\prime }=0$
Equation of other diagonal will be $2x-3y+c^{\prime }=0$ this line passes through midpoint of other diagonal i.e(2,3)

$\therefore$ we get $c^{\prime }=5$
slope of other diagonal is$\frac{2}{3}$ $⟹tan\theta =\frac{2}{3}$ $⟹cos\theta =\frac{3}{\sqrt{13}}andsin\theta =\frac{2}{\sqrt{13}}$
co-ordinates of B will be $(2-rcos\theta ,3-rsin\theta )$ and of D will be

$(2+rcos\theta ,3+rsin\theta )$
$⟹B\equiv (2-\sqrt{13}\times \frac{3}{\sqrt{13}},3-\sqrt{13}\times \frac{2}{\sqrt{13}})\equiv (-1,1)$
$⟹D\equiv (2+\sqrt{13}\times \frac{3}{\sqrt{13}},3-\sqrt{13}\times \frac{2}{\sqrt{13}})\equiv (5,5)$