QUESTION

# Find the line parallel to 2x+5y+6=0 and passing through (2, 4).

Hint: We will convert equation $2x+5y+6=0$ into slope intercept form $y=mx+c$ and then get the slope from it. Then we will substitute this slope and the coordinate (2,4) in point slope form $y-{{y}_{1}}=m(x-{{x}_{1}})$ to get our answer.

Before proceeding with the question, we should know the concepts related to the equation of lines and its different forms.
Linear equations are a combination of constants and variables.
The most common form of linear equations is slope-intercept form, which is represented as;
$y=mx+c.....(1)$ where y and x are the points in the x-y plane, m is the slope of the line (also called gradient) and c is the intercept (a constant value).
For example, $y=5x+2$. In this slope(m) is 5 and intercept is 2.
In point slope form of linear equation, a straight line equation is formed by considering the points in x-y plane, such that:
$y-{{y}_{1}}=m(x-{{x}_{1}}).......(2)$ where $({{x}_{1}},{{y}_{1}})$ are the coordinates of the line.
Now transforming $2x+5y+6=0$ in slope intercept form to find the slope of the equation. We get,
$y=-\dfrac{2}{5}x-\dfrac{6}{5}.....(3)$
Now on comparing equation (3) with equation (1) we get the slope of this line $m=-\dfrac{2}{5}$.
Now it is mentioned in the question that the equation of the line we have to find is parallel to $2x+5y+6=0$ and hence the slopes of both these equations will be equal. So now substituting the value of m in equation (2) we get,
$y-{{y}_{1}}=-\dfrac{2}{5}(x-{{x}_{1}}).......(4)$
And it is also mentioned in the question that the line passes through (2, 4). So now substituting ${{x}_{1}}=2$ and ${{y}_{1}}=4$ in equation (4) we get,
$y-4=-\dfrac{2}{5}(x-2).......(5)$
Now rearranging and simplifying equation (5) we get,
\begin{align} & \,\Rightarrow 5y-20=-2x+4 \\ & \Rightarrow 2x+5y-24=0 \\ \end{align}
Hence the line parallel to $2x+5y+6=0$ and passing through (2, 4) is $2x+5y-24=0$.

Note: An alternate and less time consuming solution is
The equation of line is $2x+5y+6=0......(6)$ So equation of any line parallel to given line can be written as $2x+5y+k=0.......(7)$
The line passes through (2,4) so substituting these coordinates in equation (7) we get,
\begin{align} & \,\Rightarrow 2\times 2+5\times 4+k=0 \\ & \,\Rightarrow k=-4-20=-24 \\ \end{align}
So the line parallel to $2x+5y+6=0$ is $2x+5y-24=0$.