Linear Functions and Graphs

Chapter 6 Study Notes · O-Level 4052

Section 6 of 9
Contents

6.5Finding the y-Intercept and Equation of a Line

Given the gradient and one point on a line, substitute into y=mx+cy = mx + c and solve for cc to get the full equation.

xy(0, 2)(2, 8)-2246810-2-112345O
Gradient 33 through (2,8)(2, 8). Reading where the line crosses the yy-axis gives c=2c = 2.

Equation from gradient and one point

A line has gradient 3\displaystyle 3 and passes through the point (2,8)\displaystyle (2, 8). Find the y\displaystyle y-intercept and write down the equation of the line.

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y\displaystyle y
=mx+c\displaystyle = mx + c
8\displaystyle 8
=3(2)+csubstitute m=3,x=2,y=8\displaystyle = 3(2) + c \quad \scriptsize\textit{substitute } m = 3, \, x = 2, \, y = 8
8\displaystyle 8
=6+c\displaystyle = 6 + c
c\displaystyle c
=2\displaystyle = 2
Equation: y=3x+2y = 3x + 2.

Full equation from two points

Find the equation of the straight line passing through C(1,1)\displaystyle C(1, -1) and D(4,8)\displaystyle D(4, 8).

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Step 1: Find the gradient:
m\displaystyle m
=8(1)41\displaystyle = \dfrac{8 - (-1)}{4 - 1}
=93\displaystyle = \dfrac{9}{3}
=3\displaystyle = 3
Step 2: Find cc using point C(1,1)C(1, -1):
1\displaystyle -1
=3(1)+c\displaystyle = 3(1) + c
c\displaystyle c
=13\displaystyle = -1 - 3
c\displaystyle c
=4\displaystyle = -4
Equation: y=3x4y = 3x - 4.
Check with D(4,8)D(4, 8): 3(4)4=83(4) - 4 = 8

Now you try

A line has gradient 4-4 and passes through (2,17)(-2,\, 17). Find the yy-intercept.

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Using y=mx+cy = mx + c with m=4m = -4 and the point (2,17)(-2,\, 17):
17\displaystyle 17
=4×(2)+c\displaystyle = -4 \times (-2) + c
17\displaystyle 17
=8+c\displaystyle = 8 + c
c\displaystyle c
=178\displaystyle = 17 - 8
c\displaystyle c
=9\displaystyle = 9

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Reverse problem: finding an unknown coordinate

A line passes through (0,5)\displaystyle (0, 5) and has gradient 3\displaystyle -3. The point (k,7)\displaystyle (k, -7) lies on the line. Find k\displaystyle k.

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The line has gradient m=3m = -3 and yy-intercept c=5c = 5.
Equation of the line: y=3x+5y = -3x + 5
Substitute the point (k,7)(k, -7):
7\displaystyle -7
=3k+5\displaystyle = -3k + 5
3k\displaystyle -3k
=12subtract 5 from both sides\displaystyle = -12 \quad \scriptsize\textit{subtract 5 from both sides}
k\displaystyle k
=4\displaystyle = 4

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