Question

For what values of k are the lines x – 2y + 1 = 0, 2x – 5y + 3 = 0 and 5x – 9y + k = 0 are concurrent. 

Answer 1

x – 2y + 1 = 0 ……. (1) 

2x – 5y + 3 = 0 ….. (2) 

5x – 9y + k = 0 …….. (3) 

Solve (1) and (2) x/(-6 + 5) = - y/(3 - 2) = 1

x/-1 = - y/-1 = 1/-1

⇒ x = 1, y = 1 

point of intersection of (1) and (2) is (1,1), 

⇒ (1, 1) lies on (3) = 5 – 9 + k = 0 = k = 4

Comments

thank you it helped