Question

Find the equations to the altitudes of the triangle whose angular points are A (2, −2), B (1, 1) and C (−1, 0).

Answer 1

Let \[m_{AD} , m_{BE} \text { and } m_{CF}\]  be the slopes of the altitudes AD, BE and CF, respectively.

\[\therefore\text {  Slope of AD } \times \text { Slope of BC } = - 1\]

\[ \Rightarrow m_{AD} \times \left( \frac{0 - 1}{- 1 - 1} \right) = - 1\]

\[ \Rightarrow m_{AD} \times \frac{1}{2} = - 1\]

\[ \Rightarrow m_{AD} = - 2\]

\[\text { Slope of BE } \times \text { Slope of AC } = - 1\]

\[ \Rightarrow m_{BE} \times \left( \frac{0 + 2}{- 1 - 2} \right) = - 1\]

\[ \Rightarrow m_{BE} \times \left( \frac{- 2}{3} \right) = - 1\]

\[ \Rightarrow m_{BE} = \frac{3}{2}\]

\[\text { Slope of CF } \times \text { Slope of AB } = - 1\]

\[ \Rightarrow m_{CF} \times \left( \frac{1 + 2}{1 - 2} \right) = - 1\]

\[ \Rightarrow m_{CF} \times \left( - 3 \right) = - 1\]

\[ \Rightarrow m_{CF} = \frac{1}{3}\]

Now, the equation of AD which passes through A (2, −2) and has slope −2 is

\[y + 2 = - 2\left( x - 2 \right)\]

\[ \Rightarrow 2x + y - 2 = 0\]

The equation of BE, which passes through B (1, 1) and has slope  \[\frac{3}{2}\] is

\[y - 1 = \frac{3}{2}\left( x - 1 \right)\]

\[ \Rightarrow 3x - 2y - 1 = 0\]

The equation of CF, which passes through C (−1, 0) and has slope  \[\frac{1}{3}\] is 

\[y - 0 = \frac{1}{3}\left( x + 1 \right)\]

\[ \Rightarrow x - 3y + 1 = 0\]