Question

How do you write the point slope of the equation given $\left( -2,4 \right)$ and parallel to $y=\dfrac{-5x}{2}+5$ ?

Answer 1

There are many different forms of only one line. There is slope intercept form, point slope, parametric equation of a line. At the end, different equations of a line talk about the same line. So, here, we are going to specifically look at the point slope form of the line. The general equation of a point slope form is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ , where $m$ is the slope of the line and $\left( {{x}_{1}},{{y}_{1}} \right)$ is a point on the line. When we say two lines or any number of lines are parallel to each other, it means they all have the same slope.

Complete step by step solution:
We are given that our line is parallel to the line $y=\dfrac{-5x}{2}+5$.
Since being parallel to a line means sharing the same slope, let us find out the slope of the given line by comparing it with the general equation of slope intercept for.
The general equation of the slope intercept form is $y=mx+c$.
Upon comparing the given line equation with the standard equation, our slope is equal to $\dfrac{-5}{2}$.
So now we have our slope$\left( m \right)$, and a point which satisfies that line, represented by $\left( {{x}_{1}},{{y}_{1}} \right)$, we are supposed to find out is already given which is $\left( -2,4 \right)$.
Let us substitute all that we got in the standard equation of slope point form of a line which is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
Upon substituting, we get the following :
$\begin{aligned} & \Rightarrow y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\\ & \Rightarrow y-4=\dfrac{-5}{2}\left( x-\left( -2 \right) \right) \\\ & \Rightarrow y-4=\dfrac{-5}{2}\left( x+2 \right) \\\ \end{aligned}$
Let us solve further to get the equation of our line.
$\begin{aligned} & \Rightarrow y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\\ & \Rightarrow y-4=\dfrac{-5}{2}\left( x-\left( -2 \right) \right) \\\ & \Rightarrow y-4=\dfrac{-5}{2}\left( x+2 \right) \\\ & \Rightarrow 2y-8=-5x-10 \\\ & \Rightarrow 5x+2y+2=0 \\\ \end{aligned}$
$\therefore$ $5x+2y+2=0$ is the slope point equation of the line which has the point $\left( -2,4 \right)$ and is parallel to the line $y=\dfrac{-5x}{2}+5$.
Graph for reference :

Note: We can either find the slope of the given line by differentiation. We have to differentiate the entire line equation with respect to $x$. It is important to remember all the general forms of a line and their standard equation so as to complete a question quickly and accurately in the exam. We should know the conditions when two lines are parallel or when two lines are said to be perpendicular. We should also be careful while substituting the values.