# The straight line L passes through the points (0, 12) and (10, 4). Find an equation of the straight line which is parallel to L and passes through the point (5, –11).? Solve without a graph paper and using graphs- show working out

Apr 23, 2018

"y=-4/5x-7

#### Explanation:

$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.

•color(white)(x)y=mx+b

$\text{where m is the slope and b the y-intercept}$

$\text{to calculate m use the "color(blue)"gradient formula}$

•color(white)(x)m=(y_2-y_1)/(x_2-x_1)

$\text{let "(x_1,y_1)=(0,12)" and } \left({x}_{2} , {y}_{2}\right) = \left(10 , 4\right)$

$\Rightarrow m = \frac{4 - 12}{10 - 0} = \frac{- 8}{10} = - \frac{4}{5}$

$\Rightarrow \text{line L has a slope } = - \frac{4}{5}$

• " Parallel lines have equal slopes"

$\Rightarrow \text{line parallel to line L also has slope } = - \frac{4}{5}$

$\Rightarrow y = - \frac{4}{5} x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$

$\text{to find b substitute "(5,-11)" into the partial equation}$

$- 11 = - 4 + b \Rightarrow b = - 11 + 4 = - 7$

$\Rightarrow y = - \frac{4}{5} x - 7 \leftarrow \textcolor{red}{\text{is equation of parallel line}}$

Apr 23, 2018

$y = - \frac{4}{5} x - 7$

#### Explanation:

First work out the gradient of L.
You can do this by using this equation- $\frac{y 1 - y 2}{x 1 - x 2}$

Lets make $\left(0 , 12\right)$ be $\left(x 1 , y 1\right)$
and $\left(10 , 4\right)$ be $\left(x 2 , y 2\right)$

Therefore the gradient is equal to- $\frac{\left(12 - 4\right)}{\left(0 - 10\right)}$

This is equal to $\frac{8}{-} 10$ or simplified $- \frac{4}{5}$.

We are now tasked with finding the equation of a line that runs parallel to L and goes through the point $\left(5 , - 11\right)$

There is a very important rule that allows us to work out the equation of parallel lines, this being that lines that are parallel all have the SAME gradient.

Therefore the new line that goes through $\left(5 , - 11\right)$ also has a gradient of $- \frac{4}{5}$ (because it is parallel)

Now as we know one point on the line and we know the gradient we can utilise the equation for a straight line- $y - y 1 = m \left(x - x 1\right)$

(where $\left(x 1 , y 1\right)$ is $\left(5 , - 11\right)$ and m is the gradient $\left(- \frac{4}{5}\right)$

Input these value and you get $y - - 11 = - \frac{4}{5} \left(x - 5\right)$

Expand and simplify and you get: $y + 11 = - \frac{4}{5} x + 4$

Put everything equal to y and you get $y = - \frac{4}{5} x - 7$

*Check this by inputting x as 5 and see if you get -11*