Question

How do you write an equation of a line given point $\left( -4,-2 \right)$ and $\left( 4,0 \right)$?

Answer 1

To write an equation of a line passing through the given points first we need to find the slope of the line by using the formula $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. Then by using the general equation formula of line $\left( y-{{y}_{2}} \right)=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{2}} \right)$ we will get the desired answer.

Complete step by step answer:
We have been given the points $\left( -4,-2 \right)$ and $\left( 4,0 \right)$.
We have to write the equation of a line passing through the given points.
Now, we know that the slope of a line passing through the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by the formula $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Now, we have ${{x}_{1}}=-4,{{y}_{1}}=-2,{{x}_{2}}=4,{{y}_{2}}=0$
Now, substituting the values in the formula we will get
$\Rightarrow \dfrac{0-\left( -2 \right)}{4-\left( -4 \right)}$
Now, simplifying the above obtained equation we will get
$\begin{aligned} & \Rightarrow \dfrac{0+2}{4+4} \\\ & \Rightarrow \dfrac{2}{8} \\\ & \Rightarrow \dfrac{1}{4} \\\ \end{aligned}$
Now, we know that the equation of the line passing through the given points is given by $\left( y-{{y}_{2}} \right)=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{2}} \right)$.
Substituting the values we will get
$\Rightarrow \left( y-0 \right)=\dfrac{1}{4}\left( x-4 \right)$
Now, simplifying the above obtained equation we will get
$\Rightarrow y=\dfrac{1}{4}x-1$
Hence above is the required equation of the line.

Note: The equation of slope-intercept form of the line is given as $y=mx+c$, where m is the slope of the line and c is the y-intercept of the line. Alternatively we can use the formula $\dfrac{y-{{y}_{1}}}{x-{{x}_{1}}}=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}$ to find the equation of a line passing through the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ without finding the slope of the line.