Question

Let $PQRS$ is a parallelogram where $P=(2,2),Q=(6,-1),\text{and}R=(7,3)$. Then equation of the line through $S$ and perpendicular to $QR$ is

• A. $x+4y=27$
• B. $4x-3y+6=0$
• C. $x+4y=10$
• D. $x+4y=2$

Answer 1

The correct option is A $x+4y=27$

$P(x_1,y_1),Q(x_2,y_2),R(x_3,y_3),\text{then}S=(x_1+x_3-x_2,y_1+y_3-y_2)$
$\therefore S=(3,6)$
Slope of $QR$ is $4$
$\therefore$ Required equation of the line is
$y-6=-\frac{1}{4}(x-3)$
$\Rightarrow x+4y=27$