Question

The verities of a quadrilateral are $A(-2,6),B(1,2),C(10,4)$ and $D(7,8)$. Find the equation of its diagonals.

• A. $6x+y+34=0,x-y-1=0$
• B. $x+6y-34=0,x-y+1=0$
• C. $x+3y-24=0,x+y-1=0$
• D. None of these

Answer 1

The correct option is B $x+6y-34=0,x-y+1=0$

Given $A(-2,6),B(1,2),C(10,4)$ and $D(7,8)$

Diagonals are $AC$ and $BD$

Midpoint of $AC(4,5)$

Midpoint of $BD(4,5)$

Slope of $AC=\frac{4-6}{10-(-2)}=\frac{-2}{12}=\frac{-1}{6}$

equation of line with slope $\frac{-1}{6}$ and passing through $(4,5)$

$y-5=\frac{-1}{6}(x-4)\Rightarrow 6y-30=-x+4\Rightarrow 6y+x-54=0$

Slope of $BD=\frac{8-2}{7-1}=\frac{6}{6}=1$

equation of line with slope $1$ and passing through $(4,5)$

$y-5=1(x-4)\Rightarrow y-x=1\Rightarrow x-y+1=0$

So equation are $x+6y-34=0$ and $x-y+1=0$