The vertices of the triangle ABC have coordinates A(-5,3) , B(-2,0) and
C(4,-1) . Find the equations of the sides of the triangle.
The equation of line joining two points
(a,b) and (c,d) is given by
$\displaystyle (y -b) = \frac{(d-b)}{(c-a)} \times{(x-a)}$
You have three points and you are asked to find the equations of those three lines which join them
Now think ahead yourself and tell if you feel trouble
The equation is the same, if you set $\displaystyle (a,\;b)=(x_1,\;y_1)$, $\displaystyle (c,\;d)=(x_2,\;y_2)$, and divide by $\displaystyle y_2 - y_1$.
You have three lines, so you will need to apply the equation three times. Just substitute the coordinates for each pair of points.
Make a sketch on the coordinate plane will make things easier .
Let me start you off with the line AB .
The gradient is -1 and we know the line passes through point (-5,3) or
(-2,0)
y-y1=m(x-x1)
thus , y=-x-2
do it this way for the remaining two sides of the triangle .