Question

find the equation of the straight line perpendicular to the line 5x+6y=1 and passes through (-1,5)

Answer 1

Answer:

ANSWER

Consider the given equation of the line.

5x−6y=1 ...........(1)

3x+2y=−5 ...........(2)

(1)+(2)×3

5x−6y=1

9x+6y=−15

14x+0=−14

x=−1 sub in (2)

3(−1)+2y=−5

2y=−2

y=−1

Point of intersection of above both lines is

=(−1,−1)

Given straight line

3x−5y+11=0

Slope of line m

1

=−

−5

3

=

5

3

Since, the line is perpendicular to the the line, so the slope the line is

m

2

=

m

1

−1

=

5

3

−1

=−

3

5

So, the equation of line passes through intersection point

y+1=−

3

5

(x+1)

3y+3=−5x−5

5x+3y+8=0

Hence, this is the answer.