Distance to the point of intersection of two altitudes in a triangle

Question

Problem

$P,Q,R$ are the points $(4,2)$, $(2,1)$ and $(6,-3)$. Also, $PS$ and $QT$ are altitudes in this triangle.

A) Find the equations of PS and QT

My answer: PS: $y-x-2=0$, QT: $5y-2x-1=0$

B) Hence find the coordinates of U, the point of intersection of PS and QT.

My answer: $(11/3, 5/3)$

C) Find the length of PU as a surd in its simplest form

This I do not know how to do.

Answer 1

HINT:

$$PS\perp QR$$ So, the gradient of PS$(m_{PS})\cdot m_{QR}=-1$

Now, $$m_{QR}=\frac{-3-1}{6-2}$$

So, the equation of $PS$ will be $$\frac{y-2}{x-4}=m_{PS}$$

Now for $(C),$ do you know the distance between $$(x_1,y_1);(x_2,y_2)$$ is $$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$