Question & Answer
QUESTION

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

ANSWER Verified Verified
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.

Complete step-by-step 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 \[2x+5y-24=0\].