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:

$x+y=1...\left(i\right)$

$2x+3y=6...\left(ii\right)$

$4x-y+4=0...\left(iii\right)$

In triangle ABC, let equations (i), (ii) and (iii) respresent the sides AB, BC and CA, respectively.

Solving (i) and (ii):

$x=-3,y=4$

Thus, AB and BC intersect at B $-3,4$ Solving (i) an (iii):

$x=-\frac{3}{5},y=\frac{8}{5}$

Thus, AB and CA intersect at A

$(-\frac{3}{5},\frac{8}{5})$

Let AD and BE be the altitudes.

$AD\perp BC$ 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 (ii)

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

$\therefore$ Slope of $AD\times (-\frac{2}{3})$ = -1 and slope of $BE\times 4=-1$

$\Rightarrow$ Slope of $AD=\frac{3}{2}$ and slope of $BE=-\frac{1}{4}$

The equation of the altitude AD passing through

$A(-\frac{3}{5},\frac{8}{5})$ and having slope $\frac{3}{2}$ is

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

$\Rightarrow 3x-2y+5=0...\left(iv\right)$

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

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

$\Rightarrow x+4y-13=0$

Solving (iv) and (v), we get $\frac{3}{7},\frac{22}{7}$ as the orthocentre of the triangle.