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

equation of first line:

y-2 = 5(x-1)
y = 5x - 5 + 2 ---> y = 5x - 3

The slope of the perpendicular must be -1/5
it meets the old line when x = 4
which means at y = 5(4)-3 or 17 and the point (4,17) must be common to both lines
new line:
y = (-1/5)x + b, but (4,17) lies on it, so ...
17 = (-1/5)(4) + b
b = 89/5

new line: y = (-1/5)x + 89/5

proof:
https://www.wolframalpha.com/input/?i=solve+y+%3D+%28-1%2F5%29x+%2B+89%2F5%2C+y+%3D+5x-3

To find the equation for line L2, we need to determine its gradient and the y-intercept.

Given that line L1 has a gradient of 5 and passes through the point (1,2), we can use the equation:

(y - y1) = m(x - x1)

where (x1, y1) is the point (1,2) and m is the gradient of L1.

Substituting the values into the equation gives:

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

Expanding and rearranging the equation:

y - 2 = 5x - 5
y = 5x - 3

So the equation for line L1 is y = 5x - 3.

Since line L2 is perpendicular to L1, its gradient will be the negative reciprocal of the gradient of L1. Therefore, the gradient of L2 is -1/5.

We know that line L2 passes through the point where X = 4. We can use the point-slope form of a line to find the equation:

(y - y1) = m(x - x1)

Substituting the values into the equation gives:

(y - y1) = (-1/5)(x - x1)

Using the point (4, y), we have:

(y - y1) = (-1/5)(x - 4)

Expanding and rearranging the equation:

y - y1 = (-1/5)x + 4/5

y = (-1/5)x + 4/5 + y1

Since we don't know the value of y1, we need to express the equation in the form y = mx + c.

To do this, we rearrange the equation:

y = (-1/5)x + y1 + 4/5

Combining the constants:

y = (-1/5)x + (y1 + 4/5)

Let c = y1 + 4/5, so the equation for L2 in the form y = mx + c is:

y = (-1/5)x + c

Therefore, the equation for L2 is y = (-1/5)x + c.

To find the equation for line L2, we first need to determine the gradient of line L1.

Given that line L1 passes through the point (1, 2) and has a gradient of 5, we can use the point-gradient form of a linear equation to find the equation for L1.

The point-gradient form of a linear equation is given by:
y - y1 = m(x - x1)

Where (x1, y1) is a point on the line, m is the gradient, and (x, y) are variables representing any point on the line.

Substituting the values of the point (1, 2) and the gradient 5 into the point-gradient form, we get:
y - 2 = 5(x - 1)

Simplifying the equation, we get:
y - 2 = 5x - 5
y = 5x - 3

So the equation for line L1 is y = 5x - 3.

Since line L2 is perpendicular to line L1, their gradients are negative reciprocals of each other. The negative reciprocal of 5 is -1/5.

To find the equation for line L2, we have a known point where the two lines intersect, which is (4, y). We need to find the value of y.

Since this point lies on both line L1 and line L2, we can substitute the x-coordinate (4) into the equation for line L1 (y = 5x - 3) and solve for y.

Substituting x = 4 into the equation for line L1, we get:
y = 5(4) - 3
y = 20 - 3
y = 17

So the point of intersection is (4, 17).

Now, using the point-gradient form for line L2, we can substitute the point of intersection (4, 17) and the gradient -1/5 into the equation to find the equation for L2.

The point-gradient form of a linear equation is:
y - y1 = m(x - x1)

Substituting the values into the equation, we get:
y - 17 = -1/5(x - 4)

Expanding and simplifying the equation, we get:
y - 17 = -1/5x + 4/5

To write the equation in the form y = mx + c, where c is the y-intercept, we rearrange the equation:
y = -1/5x + 4/5 + 17
y = -1/5x + 4/5 + 85/5
y = -1/5x + 89/5

Therefore, the equation for line L2 in the form y = mx + c is y = -1/5x + 89/5.