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Year 9AC9M9A03

Gradients — Determining the Gradient from Another Line

Find the gradient of a line segment, the midpoint of the line interval and the distance between 2 distinct points on the Cartesian plane.

Worked example

e.g. determine the gradient of the line perpendicular to y=12x+3\text{determine the gradient of the line perpendicular to } y = \tfrac{1}{2}x + 3
The given linexyy = ½x + 3
  1. Calculate the gradient of line 1m1=12m_{1} = \tfrac{1}{2}
  2. If you need parallel lines, the second line's gradient is the samem2=12m_{2} = \tfrac{1}{2}
  3. If you need perpendicular lines, use m1×m2=1m_{1} \times m_{2} = -1m1×m2=1m_{1} \times m_{2} = -1
  4. Substitute and solve for m2m_{2}12×m2=1  m2=2\tfrac{1}{2} \times m_{2} = -1 \ \therefore \ m_{2} = -2
    The original line and a perpendicular line with gradient negative twoxym₁ = ½m₂ = -2

Practice

Determine the gradient of the line:

1
parallel to y=3x1\text{parallel to } y = 3x - 1
Answerm=3m = 3
2
perpendicular to y=3x1\text{perpendicular to } y = 3x - 1
Answerm=13m = -\tfrac{1}{3}
3
parallel to 2y=3+x\text{parallel to } 2y = 3 + x
Answerm=12m = \tfrac{1}{2}
4
perpendicular to y=2x\text{perpendicular to } y = 2x
Answerm=12m = -\tfrac{1}{2}
5
perpendicular to x=y+3\text{perpendicular to } x = y + 3
Answerm=1m = -1
6
parallel to y6=x3\text{parallel to } y - 6 = x - 3
Answerm=1m = 1
7
perpendicular to x+2y=5\text{perpendicular to } x + 2y = 5
Answerm=2m = 2
8
parallel to 3y=6x9\text{parallel to } 3y = 6x - 9
Answerm=2m = 2
9
perpendicular to 4x+y=7\text{perpendicular to } 4x + y = 7
Answerm=14m = \tfrac{1}{4}
10
parallel to y=56x\text{parallel to } y = -\tfrac{5}{6}x
Answerm=56m = -\tfrac{5}{6}
Next step
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