6.5Finding the y-Intercept and Equation of a Line

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

Gradient

3

through

(2, 8)

. Reading where the line crosses the

y

-axis gives

c = 2

Equation from gradient and one point

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

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\displaystyle y

\displaystyle = mx + c

\displaystyle 8

\displaystyle = 3(2) + c \quad \scriptsize\textit{substitute } m = 3, \, x = 2, \, y = 8

\displaystyle 8

\displaystyle = 6 + c

\displaystyle c

\displaystyle = 2

Equation:

y = 3x + 2

Full equation from two points

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

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Step 1: Find the gradient:

\displaystyle m

\displaystyle = \dfrac{8 - (-1)}{4 - 1}

\displaystyle = \dfrac{9}{3}

\displaystyle = 3

Step 2: Find

c

using point

C(1, -1)

\displaystyle -1

\displaystyle = 3(1) + c

\displaystyle c

\displaystyle = -1 - 3

\displaystyle c

\displaystyle = -4

Equation:

y = 3x - 4

Check with

D(4, 8)

3(4) - 4 = 8

✓

Now you try

Find the $y$ -intercept of the line passing through $(-1,\, 3)$ and $(4,\, 23)$ .

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\displaystyle \text{Gradient}

\displaystyle = \dfrac{y_2 - y_1}{x_2 - x_1}

\displaystyle = \frac{23 - 3}{4 - (-1)}

\displaystyle = \frac{20}{5}

\displaystyle = 4

Using

y = mx + c

with

(-1,\, 3)

\displaystyle 3

\displaystyle = 4 \times (-1) + c

\displaystyle 3

\displaystyle = -4 + c

\displaystyle c

\displaystyle = 3 - (-4)

\displaystyle c

\displaystyle = 7

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Reverse problem: finding an unknown coordinate

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

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The line has gradient

m = -3

and

y

-intercept

c = 5

Equation of the line:

y = -3x + 5

Substitute the point

(k, -7)

\displaystyle -7

\displaystyle = -3k + 5

\displaystyle -3k

\displaystyle = -12 \quad \scriptsize\textit{subtract 5 from both sides}

\displaystyle k

\displaystyle = 4