Question

the point P(a,b) lies on the straight line 3x+2y=13 and the point Q(b,a) lies on the straight line 4x-y=5, then the equation of line PQ is?

Answer 1

Point P(a,b) lies on the straight line 3x + 2y = 13

so P(a,b) satisfies this straight line so, 3a + 2b = 13 (i)

and point Q(b,a) lies on the straight line 4x - y = 5

so point Q(b,a) satisfies this straight line so , 4b - a = 5 (ii)

solving equation (i) and (ii) we get

a = 3, b = 2

so P (3,2) and Q(2,3)

so the equation of line PQ will be

(y - y₁) = {(y₂ - y₁)/(x₂ - x₁)}× (x - x₁)

so (y - 2) = {(3 - 2)/(2 - 3)}× (x - 3)

so y - 2 = - x + 3

and y + x = 5 ans