Question

If a and b are two arbitrary constants, then the straight line (a-2b)x + (a+3b)y + 3a+4b = 0 will pass through

• A. (-1,-2)
• B. (1,2)
• C. (-2,-3)
• D. (2,3)

Answer 1

The correct option is A

(-1,-2)

Given equation is

(a-2b)x + (a+3b)y + 3a+4b = 0

Or a(x+y+3) + b(-2x+3y+4) = 0

This represents a family of straight lines through the point of intersection of

x + y + 3 = 0 -----------------------------( 1 )

And -2x + 3y + 4 = 0 ------------------- ( 2 )

Multiplying 2 in equation 1 and add it to equation 2

2x + 2y + 6 = 0

$\frac{-2x+3y+4=0}{5y+10=0}$

y = -2

substituting y in equation 1

x - 2 + 3 = 0

x = -1

Point of intersection of these two lines (-1,-2)

Family of straight lines should pass through (-1,-2).