The equation of line passes (-1, 4) and
perpendicular to the line 3x + 4y + 5 = 0 is
Answers
Answered by
0
Step-by-step explanation:
Slope of the given line 3x+4y+5=0 is m
1
=−
4
3
Slope of the line perpendicular to the given line is
m
1
−1
=
(
4
−3
)
−1
=
3
4
The equation of the line which is perpendicular to 3x+4y+5=0 and passing through point (4,−5) is y−y
1
=m(x−x
1
)
⇒y−(−5)=
3
4
(x−4)
⇒3y+15=4x−16
⇒4x−3y−31=0
Answered by
6
Answer:
Correct option is
A
4x−3y−31=0
Slope of the given line 3x+4y+5=0 is m
1
=−
4
3
Slope of the line perpendicular to the given line is
m
1
−1
=
(
4
−3
)
−1
=
3
4
The equation of the line which is perpendicular to 3x+4y+5=0 and passing through point (4,−5) is y−y
1
=m(x−x
1
)
⇒y−(−5)=
3
4
(x−4)
⇒3y+15=4x−16
⇒4x−3y−31=0
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