How to solve part II of this question? Given the points $A(4, 2)$ and $B(1, 1)$, find the equation of the line $l_1$ perpendicular to $AB$ passing through the point $B$. 
The point $C$ lies on $l_1$ such that $ABC$ is an isosceles triangle. Explain which two sides of the triangle must be of the same length, and find the coordinates of the two possible positions of $C$.
 A: Hints:
1) Find the slope of the line perpendicular to the line passing through the two given points.
Use this slope and point B to find the required line equation.
2)Since ABC is right angled at B, the equal sides must be 'AB' and 'BC'. So point C can be on either sides of point B with distance equal to that of AB, which is $\sqrt{10}$.
A: Draw a picture. 
Since you only ask about the second part, I assume you found that the line $l_1$ has equation $y=-3x+4$. 
Since $\angle ABC$ is a right-angle, the two equal sides must be $AB$ and $BC$. The picture will tell you at least roughly what the two possibilities for $C$ are. 
If $C$ has coordinates $(c,-3c+4)$, we can write down an equation that says the square of the distance from $C$ to $B$ is the square of the distance from $A$ to $B$, which is $10$. 
The equation is
$$(c-1)^2+(-3c+3)^2=10.$$
We get lucky when we simplify. 
A: STEP 1
find the slope of line AB $\;m_1=\dfrac {2-1}{4-1}\implies\dfrac 13$
step 2
then slope of line $l_1$which is perpendicular to $AB$ $\;\;m_2=-3$
step3
eqn of line $l_1$ which is passes from B $\;\;y-1=-3(x-1)$
$$y-1=-3x+3$$
$$3x+y=4$$
part 2
$\Delta ABC$ is a right angle triangle at B so equal sides are AB and BC.
$$AB=\sqrt{(4-1)^2+(2-1)^2}\implies\sqrt {10}=BC$$
suppose $C(x_1,y_1)$
then
$3x_1+y_1=4\implies y_1=4-3x_1\;\;$and$\;BC^2={(x_1-1)^2+(y_1-1)^2}$
$$\;10={(x_1-1)^2+(3-3x_1)^2}$$
$$\;10={x_1^2+1-2x_1+9+9x_1^2-18x_1}$$
$$10x_1^2-20x_1=0$$
$$x_1=0,2\;\;y_1=4,-2$$
so $C=(0,4)\;and\;(2,-2)$
