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. 

\[AD \perp BC \text { and }BE \perp AC\]

\[\therefore\] 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

\[\therefore \text { Slope of AD } \times \left( - \frac{2}{3} \right) = - 1 \text { and slope of BE } \times 4 = - 1\]

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

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.