
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
y17 = 1/5 (x4)
Now you can convert that to slopeintercept form.

L1:
y2 = 5(x1)
y2 = 5x5
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

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.


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.

Gradient=(Y2Y1)/(X2X1)
5=(y2)/(41)
y=17
gradient of L1×gradient of L2=1
5×L2=1
Gradient of L2=1/5
5(y17)=1(x1)
y=1/5x+17 1/5