equation of a line

**euclid2** · July 10th 2008, 09:48 AM

Determine the equation of a line with an x-intercept of 4 and passing through the point(5,2)

Am I doing it correctly?
y = m(x - p) + q
y=4(x-5)+2
y=4x-20+2
y=4x-18

Thanks

**Simplicity** · July 10th 2008, 10:11 AM

Originally Posted by euclid2

Determine the equation of a line with an x-intercept of 4 and passing through the point(5,2)

Am I doing it correctly?
y = m(x - p) + q
y=4(x-5)+2
y=4x-20+2
y=4x-18

Thanks

Explanation:
You know 2 points which are $(4, 0)$ [as you are told it passes x-axis through $x=4$ which occurs when $y=0$ ] and $(5,2)$ .

An equation of a straight line is in the form: $y=mx+c$ where $x$ and $y$ are your variables, $m$ is your gradient and $c$ is your y-intercept.

Gradient is calculated by $m = \frac{y_1-y_2}{x_1 - x_2}$ where $(x_1, y_1)$ and $(x_2, y_2)$ are two co-ordinates.

Use this gradient value and any co-ordinates to find the general form of the equation using $y-y_1 = m (x - x_1)$ where $(x_1, y_1)$ are any co-ordinates of the line.

The method and working out:
The two co-ordinates are $(4, 0)$ and $(5,2)$ .

$m = \frac{y_1-y_2}{x_1 - x_2} = \frac{2-0}{5-4} = \frac21 = 2$

$y-y_1 = m (x - x_1) \implies y-0 = 2(x-4) \implies y = 2x-8$

Check this with the other co-ordinates, $(5,2)$ .

$y = 2x-8 \implies 5 = 2(2) - 8$

Therefore the equation of the line is $y=2x-8$ .

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