Question

of the line passing through the point of Intersection of the lines x+3y-1=0, x-2y+4=0 and perpendicular to the line 3x+2y=0​

Answer 1

EXPLANATION.

=> equation of line passing through the point

of intersection of line

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

=> x - 2y + 4 = 0 ......(2)

perpendicular to the line = 3x + 2y = 0

$\sf : \implies \: x \: + 3y - 1 = 0 \: \: ......(1) \\ \\ \sf : \implies \: \: x \: - 2y + 4 = 0 \: \: ....(2) \\ \\ \sf : \implies \: \: we \: get \\ \\ \sf : \implies \: \: 5y - 5 = 0 \\ \\ \sf : \implies \: \: y = 1$

$\sf : \implies \: \: put \: the \: value \: of \: y \: = 1 \: in \: equation \: (1) \\ \\ \sf : \implies \: \: x + 3(1) - 1 = 0 \\ \\ \sf : \implies \: \: x \: = - 2$

$\sf : \implies \: \: point \: of \: intersection \: on \: the \: line \: ( - 2 , 1) \\ \\ \sf : \implies \: equation \: of \: line \: perpendicular \: to \: 3x + 2y = 0 \\ \\ \sf : \implies \: 2x - 3y = \lambda \: \: \: .......(3) \\ \\ \sf : \implies \: put \: the \: value \: in \: equation \: \\ \\ \sf : \implies 2( - 2) - 3(1) = \lambda \\ \\ \sf : \implies \: - 4 - 3 = \lambda \: \\ \\ \sf : \implies \: \lambda \: = - 7$

$\sf : \implies \: \therefore \: fron \: equation \: (3) \\ \\ \sf : \implies \: 2x - 3y = - 7 \\ \\ \sf : \implies \: 2x - 3y + 7 = 0$

Answer 2

Any line passing through the intersection of x+3y-1 = 0 and x-2y+4 = 0 has equation of the form x+3y-1 + k (x-2y+4) = 0.

i.e., (1+k)x+(3–2k)y -1+4k =0. ……..(1)

If this perpendicular to 3x+2y=0 then the product of their slopes = -1.

Hence,{-(1+k)/(3–2k)}{-3/2)} = -1. Solving we find that k = 9. Substituting this value of k in (1), we get the equation of the required line as 10x-15y+35 = 0.