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

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

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=(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

To find the equation for line L2, we need to first determine its gradient. Given that L2 is perpendicular to L1, we know that the product of their gradients will be -1.

The gradient of L1 is 5, so we can calculate the gradient of L2 by taking the negative reciprocal of 5:

Gradient of L2 = -1/5

We also know that L2 passes through the point where x = 4. Let's call this point (4, y).

Now we can use the point-slope form of a linear equation (y - y1) = m(x - x1) to find the equation for L2:

(y - 2) = (-1/5)(x - 4)

Next, let's simplify the equation by distributing -1/5 to (x - 4):

y - 2 = (-1/5)x + (4/5)

To isolate y, let's add 2 to both sides of the equation:

y = (-1/5)x + (4/5) + 2

Simplifying further:

y = (-1/5)x + (4/5) + 10/5

y = (-1/5)x + (14/5)

Therefore, the equation for line L2 in the form of y = mx + c is:

y = (-1/5)x + 14/5

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.