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
Since, the line is perpendicular to the the line, so the slope the line is
m
2
−1
So, the equation of line passes through intersection point
y+1=−
(x+1)
3y+3=−5x−5
5x+3y+8=0
Please Mark me BRAINLIEAST
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Answers & Comments
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
Please Mark me BRAINLIEAST
5x-11y= x = 31/5
y= 31/11 x 5