Question

Find the equations of the sides of the triangles the coordinates of whose angular points are respectively
(i) (1, 4), (2, −3) and (−1, −2)
(ii) (0, 1), (2, 0) and (−1, −2).

Answer 1

(i) Let the given points be A (1, 4), B (2, −3) and C (−1, −2).

Let $m_1,m_2\mathrm{and}{m}_3$ be the slopes of the sides AB, BC and CA, respectively.

$$\begin{gathered} \therefore m_1=\frac{-3-4}{2-1},m_2=\frac{-2+3}{-1-2}\mathrm{and}{m}_3=\frac{4+2}{1+1} \\ \Rightarrow m_1=-7,m_2=-\frac{1}{3}\mathrm{and}{m}_3=3 \end{gathered}$$

So, the equations of the sides AB, BC and CA are

$$\begin{gathered} y-4=-7\left(x-1\right),y+3=-\frac{1}{3}\left(x-2\right)\mathrm{and}y+2=3\left(x+1\right) \\ \Rightarrow 7x+y=11,x+3y+7=0\mathrm{and}3x-y+1=0 \end{gathered}$$

(ii) Let the given points be A (0, 1), B (2, 0) and C (−1, −2).

Let $m_1,m_2\mathrm{and}{m}_3$ be the slopes of the sides AB, BC and CA, respectively.

$$\begin{gathered} \therefore m_1=\frac{0-1}{2-0},m_2=\frac{-2-0}{-1-2}\mathrm{and}{m}_3=\frac{1+2}{0+1} \\ \Rightarrow m_1=-\frac{1}{2},m_2=\frac{2}{3}\mathrm{and}{m}_3=3 \end{gathered}$$

So, the equations of the sides AB, BC and CA are

$$\begin{gathered} y-1=-\frac{1}{2}\left(x-0\right),y-0=\frac{2}{3}\left(x-2\right)\mathrm{and}y+2=3\left(x+1\right) \\ \Rightarrow x+2y=2,2x-3y=4\mathrm{and}3x-y+1=0 \end{gathered}$$