How do you write an equation of a line through (-3,1) and perpendicular to y= -3x+7?

Question

Abigail Allen · Accepted Answer

#y=1/3x+2# The gradient of #y=-3x+7# is -3 The gradient of a line perpendicular is #1/3# To prove this to yourself draw a line with gradient -3 on squared paper and the draw a line at right angles So the line of the perpendicular will be in the form #y=1/3x +c# But we know it goes through (-3,1) So substituting those values gives #1=1/3*-3+c# 1=-1+c c=2

Eva Adamson · Answer

undefined To write the equation of a line perpendicular to a given line, first find the negative reciprocal of the slope of the given line. Then, use the point-slope form of the equation of a line with one of the given points.

Given line: y = -3x + 7
Slope of the given line: -3
Negative reciprocal of the slope: 1/3

Using the point (-3, 1) and the negative reciprocal slope 1/3:
y - 1 = (1/3)(x + 3)

This is the equation of the line passing through (-3, 1) and perpendicular to y = -3x + 7.