Question

Find the equation of the side BC of the triangle ABC whose vertices are A (-1, -2), B (0, 1) and C (2, 0) respectively. Also, find the equation of the median through A (-1, -2).

Answer 1

$\text{Equation of BC}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)y-1=\frac{0-1}{2-0}(x-0)[\because B(0,1),C(2,0\left)\right]2y-2=-xx+2y=2\Rightarrow x+2y-2=0\text{D is mid point of BC so}D=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{0+2}{2},\frac{1+0}{2})=(1,\frac{1}{2})\therefore \text{Equation of the median AD :}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)y-(-2)=\frac{\frac{1}{2}-(-2)}{1-(-1)}(x-(-1)=\frac{\frac{5}{2}}{2}(x+1\left)\right)[\because A(-1,-2),D(1,\frac{1}{2})]4y+8=5x+55x-4y-3=0$