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

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

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Ava 3 months 2021-07-17T14:11:50+00:00 1 Answers 0 views 0

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    2021-07-17T14:13:16+00:00

    Answer:

    Equation of the given line: 3x + 4y -5 = 0

    By rearranging, we get 4y = -3x + 5

    or y = (-3/4)x + 5/4….(1)

    The standard equation of a straight line with slope m and y-intercept c is given by :

    y = mx + c….(2)

    Comparing (1) and (2), we get m = -3/4 and c = 5/4.

    Since the required line is perpendicular to (1), its slope M is -(1/m) = 4/3

    We know that the required line passes through the point (5,1)

    The equation of this line is thus given by y-y1 =M(x-x1); where x1 is 5, y1 is 1 and M is 4/3

    Substituting, we get y-1 = 4/3(x-5)

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