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.

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

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

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

$$\Rightarrow m_{AD}=-2$$

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

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

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

$$\Rightarrow m_{BE}=\frac{3}{2}$$

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

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

$$\Rightarrow m_{CF}\times (-3)=-1$$

$$\Rightarrow m_{CF}=\frac{1}{3}$$

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

$$y+2=-2(x-2)$$

$$\Rightarrow 2x+y-2=0$$

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

$$y-1=\frac{3}{2}(x-1)$$

$$\Rightarrow 3x-2y-1=0$$

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

$$y-0=\frac{1}{3}(x+1)$$

$$\Rightarrow x-3y+1=0$$