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GB-23-04-001 · Perpendicular lines

Find the equation of the line perpendicular to y=3x+8y=3x + 8 through (3,−6)(3,-6).

Coordinate Geometry: Perpendicular Lines Booklet p. 81–82
Method
1) Identify the original gradient
2) Take the negative reciprocal
3) Substitute into y−y1=m(x−x1)y-y_1=m(x-x_1)
4) Clear any denominator before expanding and rearrange
Worked Example (GB-23-04-001)
1)
The original gradient is 33. Use the negative reciprocal: m=−13=−13m=-\frac{1}{3}=-\frac{1}{3}.
2)
Use the booklet's point-gradient formula: y−y1=m(x−x1)y-y_1=m(x-x_1).
3)
y−(−6)=−13(x−(3))y-(-6)=-\frac{1}{3}(x-(3))
4)
Clear the denominator before expanding: 3(y−(−6))=−1(x−(3))3(y-(-6))=-1(x-(3)).
5)
Expand: 3y+18=−x+33y + 18=-x + 3.
6)
Rearrange: y=−13x−5y=-\frac{1}{3}x - 5.
Ans = y=−13x−5y=-\frac{1}{3}x - 5