Question

The equation of line passes (-1, 4) and
perpendicular to the line 3x + 4y + 5 = 0 is​

Answer 1

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

Answer 2

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