Question

The two lines y=2x+8 and y=2x-12 intersect the x axis at P and Q. Work out the distance PQ

Answer 1

$\textbf{Given lines are}$

$y=2x+8\;\&\;y=2x-12$

$\text{Put y=0, in both equations, we get}$

$0=2x+8\;\&\;0=2x-12$

$2x=-8\;\&\;2x=12$

$x=-4\;\&\;x=6$

$\therefore\text{The given lines meeet x-axis at P(-4,0) and Q(6,0)}$

$\text{Now}$

$\text{The distance between P and Q is}$

$PQ= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$PQ= \sqrt{(-4-6)^2+(0-0)^2}$

$PQ= \sqrt{100}$

$\implies\boxed{\bf\,PQ= 10}$