Question

Find the equation of straight line passing through the points (-8, 2) as vertically an
angle 45° with straight line 3x - y+6=0​

Answer 1

Answer:

Explanation: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.