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Question

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

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Solution

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 = -35 , y = 85

Thus, AB and CA intersect at A -35, 85.

Let AD and BE be the altitudes.

ADBC and BEAC

Slope of AD × Slope of BC = −1
and Slope of BE × Slope of AC = −1

Here, slope of BC = slope of the line (2) = -23 and slope of AC = slope of the line (3) = 4

Slope of AD×-23=-1 and slope of BE×4=-1Slope of AD=32 and slope of BE=-14

The equation of the altitude AD passing through A -35, 85 and having slope 32 is

y-85=32x+35

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

The equation of the altitude BE passing through B (−3, 4) and having slope -14 is

y-4=-14x+3

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

Solving (4) and (5), we get 37, 227 as the orthocentre of the triangle.

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