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

Question

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

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Arya 10 months 2021-07-24T05:31:58+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-24T05:33:06+00:00

    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

    0
    2021-07-24T05:33:49+00:00

    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

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