The equation of line passes (-1, 4) andperpendicular to the line 3x + 4y + 5 = 0 is About the author Arya
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 Reply
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 Reply
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
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