Question

Find the orthocentre of the triangle the equations of whose sides are x + y = 1, 2x + 3y = 6 and 4x − y + 4 = 0.

Answer 1

The given lines are as follows:

x + y = 1 ... (1)

2x + 3y = 6 ... (2)

4x − y + 4 = 0 ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.


Solving (1) and (2):
x = −3, y = 4

Thus, AB and BC intersect at B (−3, 4).

Solving (1) and (3):
x = $-\frac{3}{5}$ , y = $\frac{8}{5}$

Thus, AB and CA intersect at $A\left(-\frac{3}{5},\frac{8}{5}\right)$.

Let AD and BE be the altitudes.

$AD\perp BC\mathrm{and}BE\perp AC$

$\therefore$ Slope of AD $\times$ Slope of BC = −1
and Slope of BE $\times$ Slope of AC = −1

Here, slope of BC = slope of the line (2) = $-\frac{2}{3}$ and slope of AC = slope of the line (3) = 4

$$\begin{gathered} \therefore \mathrm{Slope}\mathrm{of}AD\times \left(-\frac{2}{3}\right)=-1\mathrm{and}\mathrm{slope}\mathrm{of}BE\times 4=-1 \\ \Rightarrow \mathrm{Slope}\mathrm{of}AD=\frac{3}{2}\mathrm{and}\mathrm{slope}\mathrm{of}BE=-\frac{1}{4} \end{gathered}$$

The equation of the altitude AD passing through $A\left(-\frac{3}{5},\frac{8}{5}\right)$ and having slope $\frac{3}{2}$ is

$y-\frac{8}{5}=\frac{3}{2}\left(x+\frac{3}{5}\right)$

$\Rightarrow 3x-2y+5=0$ ... (4)

The equation of the altitude BE passing through B (−3, 4) and having slope $-\frac{1}{4}$ is

$y-4=-\frac{1}{4}\left(x+3\right)$

$\Rightarrow x+4y-13=0$ ... (5)

Solving (4) and (5), we get $\left(\frac{3}{7},\frac{22}{7}\right)$ as the orthocentre of the triangle.