# 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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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).
(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=(y-2)/(4-1)
y=17

5×L2=-1