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A line L1 passes through point (1, 2) and has gradient of 5. Another line L2, is perpendicular to L2 and meets it at a point where x = 4. Find the equation for L2 in the form of y = mx + c

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5 answers
  1. On L1, as x increases by 3, y increases by 5*3, so (4,17) is on L1

    The perpendicular line has slope -1/5, so L2 is

    y-17 = -1/5 (x-4)

    Now you can convert that to slope-intercept form.

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  2. L1:
    y-2 = 5(x-1)
    y-2 = 5x-5
    5x - y = 3

    L2: x + 5y = c

    from L1, when x = 4, y = 17
    so (4,17) into L2
    4 + 85 = c = 89

    L2: x + 5y = 89 or y = (-1/5)x + 89/5

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  3. The line l1 has equation   3x + 5y − 7 = 0
    (a)  Find the gradient of l1
    (2)
    The line l2 is perpendicular to l1 and passes through the point (6, −2).
    (b)  Find the equation of l2 in the form y = mx + c, where m and c are constants.

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  4. The point A has coordinates (−4, 11) and the point B has coordinates (8, 2).
    (a)  Find the gradient of the line AB, giving your answer as a fully simplified fraction.
    (2)
    The point M is the midpoint of AB. The line l passes through M and is perpendicular to AB.
    (b)  Find an equation for l, giving your answer in the form px + qy + r = 0 where p, q and r are integers to be found.
    (4)
    The point C lies on l such that the area of triangle ABC is 37.5 square units.
    (c)  Find the two possible pairs of coordinates of point C.

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  5. Gradient=(Y2-Y1)/(X2-X1)
    5=(y-2)/(4-1)
    y=17

    gradient of L1×gradient of L2=-1
    5×L2=-1
    Gradient of L2=-1/5

    5(y-17)=-1(x-1)
    y=-1/5x+17 1/5

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