Question

The orthocentre of the triangle formed by the lines 4x – 7y + 10 = 0, x + y = 5 and 7x + 4y = 15, is ______.

Options

• (1, 2)

• (1, –2)

• (–1, –2)

• (–1, 2)

Answer 1

The orthocentre of the triangle formed by the lines 4x – 7y + 10 = 0, x + y = 5 and 7x + 4y = 15, is (1, 2).

Explanation:

Given lines formed a right-angled triangle and we know that in a right-angled triangle, the orthocenter is that point where the right angle is formed. Here 4x – 7y + 10 = 0 and 7x + 4y = 15 are the lines which formed the right angle.

The point of intersection of these lines is the orthocentre

4x – 7y + 10 = 0 ...(1)

7x + 4y – 15 = 0  ...(2)

Multiply (1) by 7 and (2) by 4 we get

28x − 49y + 70 = 0
28x + 16y − 60 = 0
−     −        +            
        −65y = −130

y = `130/65`

y = 2

From (1) we get

4x – 14 + 10 = 0

⇒ 4x = 4

⇒ x = 1

∴ Orthocentre is (1, 2)