Question

How do you find the slope of the line passing through the points $\left( -7,3 \right)$ and $\left( 3,8 \right)$ ?

Answer 1

In this question we need to find the slope of the line passing through the points $\left( -7,3 \right)$ and$\left( 3,8 \right)$ . For this we will use the formulae for finding the slope of the line passing through $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given as $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .

Complete step-by-step solution:
Now considering from the question we have been asked to find the slope of the line passing through the points $\left( -7,3 \right)$ and $\left( 3,8 \right)$ .
From the basic concepts of straight line we know the formulae for finding the slope of the line passing through $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given as $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
By applying this formula in this question we will have the slope of the given straight line as
$\begin{aligned} & \Rightarrow \dfrac{8-3}{3-\left( -7 \right)} \\\ & \Rightarrow \dfrac{5}{10} \\\ & \Rightarrow \dfrac{1}{2} \\\ \end{aligned}$
Therefore we can conclude that the slope of the line passing through the points $\left( -7,3 \right)$ and $\left( 3,8 \right)$ is $\dfrac{1}{2}$.

Note: While answering questions of this type we should be sure with our concepts and the calculations we make. If we are aware of the formulae then it is a very easy and very time efficient solution. Similarly by extending this we can find the equation of the line passing through the points $\left( -7,3 \right)$ and$\left( 3,8 \right)$ given by the formulae $\left( y-{{y}_{2}} \right)=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{2}} \right)$ by using this we will have
$\begin{aligned} & \Rightarrow \left( y-8 \right)=\left( \dfrac{1}{2} \right)\left( x-3 \right) \\\ & \Rightarrow 2\left( y-8 \right)=x-3 \\\ & \Rightarrow 2y-16=x-3 \\\ & \Rightarrow x-2y+16-3=0 \\\ & \Rightarrow x-2y+13=0 \\\ \end{aligned}$ .
From the basic concept we know that the slope-intercept form of the equation is given as $y=mx+c$ where $m$ is the slope and $c$ is the constant.