Question

Find the coordinates of the orthocentre of the triangles whose angular points are $(1,0),(2,-4)$, and $(-5,-2)$.

Answer 1

Given,

Let $A(-5,-2)$, $B(1,0)$ and $C(2,-4)$be the vertices of triangle $ABC.$

Step 1 Find the equation of $CE$ and $AB$

To find the slope of AB

$$\begin{gathered} =\frac{(y_2-y_1)}{(x_2-x_1)} \\ =\frac{(0+2)}{(1+5)} \\ =\frac{2}{6} \\ =\frac{1}{3} \end{gathered}$$

Since, when lines are perpendicular the slope of the other line is $\frac{-1}{m}$

Then

Slope of $CE=-3$

The equation of $CE$ is $y+4=-3(x-2)$

$$\begin{gathered} \Rightarrow y=-3x+6–4 \\ \Rightarrow y=-3x+2..\left(i\right) \end{gathered}$$

Step 2 Finding the slope of $BC$ and $AD$

To find the slope of BC

$$\begin{gathered} =\frac{(y_2-y_{1_{}})}{(x_{2_{}}-x_1)} \\ =\frac{(-4+0)}{(2-1)} \\ =\frac{-4}{1} \\ =-4 \end{gathered}$$

Slope of $BC=-4$

The slope of $AD=\frac{1}{4}$

Equation of AD

$\begin{array}{rcl}& \Rightarrow & y+2=\frac{1}{4}(x+5)\end{array}$

$$\begin{gathered} 4y+8=x+5 \\ 4y=x–3..\left(ii\right) \end{gathered}$$

Step 3 Solve the equations to find orthocenter

Orthocenter is the point of intersection of the perpendicular from opposite vertices.

So it is the point of intersection of $AD$ and $CE$

By solving (i) and (ii) we get

$$\begin{gathered} x=\frac{11}{13}, \\ y=\frac{-7}{13} \end{gathered}$$

Hence the orthocenter is $(11/13,-7/13).$