Question

What is the equation of the line with a slope $m=\dfrac{3}{8}$ that passes through $\left( -7,-3 \right)$?

Answer 1

When slope and point are given and we need to find out the equation of all line then always remember the point-slope equation form of line i.e. $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. Here we just need to put the value and simplify it, so we will get the equation. Other than this there are other types of form also there for finding out the equation of line based on the given information.

Complete step by step solution:
As we have given the slope and point of the equation i.e. $m=\dfrac{3}{8}$and $\left( -7,-3 \right)$respectively. BY comparing it with the point-slope form of a straight line$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$, so $m=\dfrac{3}{8}$, and $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -7,-3 \right)$.
Now to find out the equation simply put the value in point-slope form.
So,

& y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\\ & y-\left( -3 \right)=\dfrac{3}{8}\left( x-\left( -7 \right) \right) \\\ \end{aligned}$$ Now, we will open brackets further and simplify them. Therefore, we get $$\Rightarrow y+3=\dfrac{3}{8}\left( x+7 \right)$$ Now we will cross multiply the terms. Therefore, we get $$\Rightarrow 8\left( y+3 \right)=3\left( x+7 \right)$$ Now, when we will open brackets, we will get $$\Rightarrow 8y+24=3x+21$$ Now, we will take constant terms on one side and variable terms on the other side, we will get $$\begin{aligned} & \Rightarrow 24-21=3x-8y \\\ & \Rightarrow 3=3x-8y \\\ \end{aligned}$$ Here we got the equation i.e. $$3=3x-8y$$. **Note:** To find the Equation of straight line we just need to have any two things, it can be a combination of two points of the straight line, or point or slope, or a two intercept form. All of these conditions have straight-line equation form. We just need to figure out these values and put in these form. Other than point-slope form $\dfrac{x}{a}+\dfrac{y}{b}=1$ is an intercept form where a and b are intercepted on x and y-axis respectively. And $$y-{{y}_{1}}=\dfrac{\left( y{}_{2}-{{y}_{1}} \right)}{\left( x{}_{2}-{{x}_{1}} \right)}\left( x-{{x}_{1}} \right)$$ is a two-point form of the equation were $\left( {{x}_{1}},{{y}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}} \right)$are two points of a straight line.