The foot of the perpendicular

Question

Find, in terms of $a$ and $b$, the coordinates of the foot of the perpendicular from the point $(a,b)$ to the line $x+2y-4=0$

I've graphed this and found the point of intersection, but not sure how to get the coordinates in terms of $a$ and $b$. Can someone help me please? Thank you.

Answer 1

Hint: Write the given equation in the form

$$y=-\frac{1}{2}x+2$$ then the slope of the perpendicular equation is given by $$y=2x+n$$ For the given Point we get

$$b=2a+n$$ thus we obtain: $$y=2x+b-a$$ Now you have to solve the equation

$$2x+b-2a=-\frac{1}{2}x+2$$ for $$x$$ For consideration: $$x=\frac{2}{5}(2+2a-b)$$

Answer 2

HINT

We can solve as follow

• consider the line through $(a,b)$ and perpendicular to $x+2y-4=0$, that is

$$(y-b)=-\frac12(x-a)$$

• find the intersection point between the two lines, that is solve the system

$$\begin{cases}x+2y-4=0\\(y-b)=-\frac12(x-a)\end{cases}$$

Answer 3

Any point$(P)$ on $x+2y-4=0$ can be set as $(4-2a,a)$

Now the gradient of $(a,b)$ and $P$ will be $\dfrac{-1}{-\dfrac12}$