Question

The orthocentre of the triangle formed by the lines x + y = 1, 2x + 3y = 6 and 4x – y + 4 = 0 lies in-

• A. 1^st quadrant
• B. 2^nd quadrant
• C. 3^rd quadrant
• D. 4^th quadrant

Answer 1

Answer: (A)

1^st quadrant

Answer 2

Intersection point of x + y = 1 and 2x + 3y = 6 is (–3, 4)

Altitude through (–3, 4) and perpendicular to the line

4x – y + 4 = 0 is

x +4y – 13 = 0 … (1)

Intersection point of x + y = 1 and 4x – y + 4 = 0 is $\left( \frac { - 3 } { 5 } , \frac { 8 } { 5 } \right)$

Altitude through $\left( \frac { - 3 } { 5 } , \frac { 8 } { 5 } \right)$and perpendicular to the line

2x + 3y = 6 is

3x – 2y + 5 = 0 … (2)

Solving equations (1) and (2), we have

H = $\left( \frac { 3 } { 7 } , \frac { 22 } { 7 } \right)$

which lies in the first quadrant.