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

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

Explanation

Solution

Hint: We will convert equation 2x+5y+6=02x+5y+6=0 into slope intercept form y=mx+cy=mx+c and then get the slope from it. Then we will substitute this slope and the coordinate (2,4) in point slope form yy1=m(xx1)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)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+2y=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:
yy1=m(xx1).......(2)y-{{y}_{1}}=m(x-{{x}_{1}}).......(2) where (x1,y1)({{x}_{1}},{{y}_{1}}) are the coordinates of the line.
Now transforming 2x+5y+6=02x+5y+6=0 in slope intercept form to find the slope of the equation. We get,
y=25x65.....(3)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=25m=-\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=02x+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,
yy1=25(xx1).......(4)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 x1=2{{x}_{1}}=2 and y1=4{{y}_{1}}=4 in equation (4) we get,
y4=25(x2).......(5)y-4=-\dfrac{2}{5}(x-2).......(5)
Now rearranging and simplifying equation (5) we get,

& \,\Rightarrow 5y-20=-2x+4 \\\ & \Rightarrow 2x+5y-24=0 \\\ \end{aligned}$$ Hence the line parallel to $$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{aligned} & \,\Rightarrow 2\times 2+5\times 4+k=0 \\\ & \,\Rightarrow k=-4-20=-24 \\\ \end{aligned}$$ So the line parallel to $$2x+5y+6=0$$ is $$2x+5y-24=0$$.