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

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Last updated date: 13th Jul 2024
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Answer
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Hint: 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
$\begin{align}
  & \Rightarrow \dfrac{8-3}{3-\left( -7 \right)} \\
 & \Rightarrow \dfrac{5}{10} \\
 & \Rightarrow \dfrac{1}{2} \\
\end{align}$
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
$\begin{align}
  & \Rightarrow \left( y-8 \right)=\left( \dfrac{1}{2} \right)\left( x-3 \right) \\
 & \Rightarrow 2\left( y-8 \right)=x-3 \\
 & \Rightarrow 2y-16=x-3 \\
 & \Rightarrow x-2y+16-3=0 \\
 & \Rightarrow x-2y+13=0 \\
\end{align}$ .
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.