A second line L2 is perpendicular to L1 at R . Find the equation of L2 in the form y=mx+c. Where m and c are constants

Answer 1

Since line L2 is perpendicular to line L1, the slope of L2 is the negative reciprocal of the slope of L1.

Let's say the equation of L1 is y = m1x + c1. The slope of L1 is m1.

So, the slope of L2, m2, is given by m2 = -1/m1.

To find the equation of L2, we need the slope and a point on the line. We know that L2 is perpendicular to L1 at point R.

Let's say the coordinates of point R are (xR, yR).

Using the point-slope form of a line, the equation of L2 can be written as:

y - yR = m2(x - xR)

Substituting the value of m2, we get:

y - yR = (-1/m1)(x - xR)

Rearranging the equation, we get:

y = (-1/m1)x + (xR/m1) + yR

Comparing this equation with the standard form y = mx + c, we get:

m = -1/m1

c = (xR/m1) + yR

So, the equation of L2 in the form y = mx + c is:

y = (-1/m1)x + (xR/m1) + yR, where m = -1/m1 and c = (xR/m1) + yR.