Question

Equation of a line passing through the point of intersection of lines 2x−3y+4=02x−3y+4=0, 3x+4y−5=03x+4y−5=0 and perpendicular to 6x−7y+3=06x−7y+3=0, then its equation is

Answer 1

first, find point of intersection of 2x - 3y + 4 = 0 and 3x + 4y - 5 = 0.

2x - 3y + 4 = 0 .........(1) × 4

3x + 4y - 5 = 0 .........(2) × 3

...........................

8x - 12y + 16 + 9x + 12y - 15 = 0

17x + 1 = 0, x = -1/17

and 4y = 5 - 3(-1/17) = 5 + 3/17

4y = 88/17, y = 22/17

hence, point of intersection is (-1/17, 22/17)

a/c to question,

line is perpendicular to 6x - 7y + 3 = 0

so, slope of line = 6/7

so, slope of unknown line × slope of {6x - 7y + 3 = 0 } = -1/(6/7) = -7/6

now, equation of unknown line is given by $(y-y_1)=m(x-x_1)$

where , $(x_1,y_1)$ = (-1/17, 22/17) and m = -7/6

e.g., (y - 22/17) = -7/6(x + 1/17)

or, (y - 22/17) + 7/6(x + 1/17) = 0

or, (17y - 22) + 7/6(17x + 1) = 0

or, 6(17y - 22) + 7(17x + 1) = 0

or, 102y - 132 + 119x + 7 = 0

hence, 119x + 102y - 125 = 0 is required equation of line.