Asked by Dydra
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
Answers
Answered by
GPT 3.5
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.
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.