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 $\displaystyle (4, 0)$ [as you are told it passes x-axis through $\displaystyle x=4$ which occurs when $\displaystyle y=0$] and $\displaystyle (5,2)$.
An equation of a straight line is in the form: $\displaystyle y=mx+c$ where $\displaystyle x$ and $\displaystyle y$ are your variables, $\displaystyle m$ is your gradient and $\displaystyle c$ is your y-intercept.
Gradient is calculated by $\displaystyle m = \frac{y_1-y_2}{x_1 - x_2}$ where $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$ are two co-ordinates.
Use this gradient value and any co-ordinates to find the general form of the equation using $\displaystyle y-y_1 = m (x - x_1)$ where $\displaystyle (x_1, y_1)$ are any co-ordinates of the line.
The method and working out:
The two co-ordinates are $\displaystyle (4, 0)$ and $\displaystyle (5,2)$.
$\displaystyle m = \frac{y_1-y_2}{x_1 - x_2} = \frac{2-0}{5-4} = \frac21 = 2$
$\displaystyle y-y_1 = m (x - x_1) \implies y-0 = 2(x-4) \implies y = 2x-8$
Check this with the other co-ordinates, $\displaystyle (5,2)$.
$\displaystyle y = 2x-8 \implies 5 = 2(2) - 8 $
Therefore the equation of the line is $\displaystyle y=2x-8$.