Question

Given that FQ perpendicularly bisects ED, solve for x and y. Need a full explanation.

Answer 1

well, we know the angle is a right-angle at Q since FQ ⟂ ED, that can only happen if FD = FE, so we have an isosceles triangle, so then

`\stackrel{EF}{x^2-5}~~ = ~~\stackrel{DF}{7x-17}\implies x^2-7x-5=-17\implies x^2-7x+12=0 \\\\\\ (x-4)(x-3)=0\implies x= \begin{cases} 3\\ 4 \end{cases}`

we also know the angle at Q is 90°, so then

`3y^2+12~~ = ~~90\implies 3y^2=78\implies y^2=\cfrac{78}{3} \\\\\\ y^2=26\implies y=√(26)\implies y\approx 5.01`