Question

Find the equation of the line through point (−2,−1) and perpendicular to 5x+6y=−6

Answer 1

Explanation:

eq. of any line perpendicular to 5x+6y=-6 is

5y-6x=c

where c is a constant.

it passes through (-2,-1) so

5(-1)-6(-2)=c

c=-5+12=7

so reqd. eq. is 5y-6x=7

or 6x-5y=-7

or

slope of given line =-5/6

slope of reqd. line=6/5

eq. of line through (-2,-1) with slope 6/5 is

y+1=6/5(x+2)

5y+5=6x+12

5y=6x+12-5

or 5y=6x+7

or

6x-5y=-7