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Question: How do you write the equation given \(\left( -6,7 \right)\); parallel to \(3x+7y=3\)?...

How do you write the equation given (6,7)\left( -6,7 \right); parallel to 3x+7y=33x+7y=3?

Explanation

Solution

In this problem we need to calculate the equation of the line which is parallel to the given line and passing through the given point. For this problem we are going to use the geometry rule that two parallel lines should have equal slope. So, we will consider the slope of the given line as m1{{m}_{1}} and the slope of the required line as m2{{m}_{2}}. We know that the slope of the line which is in the form of ax+by+c=0ax+by+c=0 is ab-\dfrac{a}{b}. From this we can have the value of m1{{m}_{1}} as well as m2{{m}_{2}}. Now we have the slope of the required line a point through which the line passes, so we will use the slope point formula and substitute all the values we have and simplify the equation to get the required solution.

Complete step-by-step solution:
Given line is 3x+7y=33x+7y=3.
Comparing the above equation with stand equation which is ax+by+c=0ax+by+c=0, then we will get
a=3a=3, b=7b=7, c=3c=-3.
Let us assume the slope of the given line as m1{{m}_{1}} and slope of the required line as m2{{m}_{2}}.
Now the slope of the given line is calculated by
m1=ab m=37 \begin{aligned} & \Rightarrow {{m}_{1}}=-\dfrac{a}{b} \\\ & \Rightarrow m=-\dfrac{3}{7} \\\ \end{aligned}
In geometry we can say that the slopes of two parallel lines are equal to each other. So, the slope of the required line will be m2=m1=37{{m}_{2}}={{m}_{1}}=-\dfrac{3}{7}. Now we have the slope of the required line and it passes through the point (6,7)\left( -6,7 \right). From slope point equation we are going to write the equation of the line as
yy1=m(xx1) y7=37(x(6)) 7(y7)=3(x+6) 7y49=3x18 3x+7y=31 \begin{aligned} & \Rightarrow y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\\ & \Rightarrow y-7=-\dfrac{3}{7}\left( x-\left( -6 \right) \right) \\\ & \Rightarrow 7\left( y-7 \right)=-3\left( x+6 \right) \\\ & \Rightarrow 7y-49=-3x-18 \\\ & \Rightarrow 3x+7y=31 \\\ \end{aligned}
Hence the equation of the required line is 3x+7y=313x+7y=31.

Note: We can follow another simple method to solve this problem. In geometry we can write the equation of the line which is parallel to the line ax+by=cax+by=c as ax+by=dax+by=d. We will substitute the point we have in the equation ax+by=dax+by=d and calculate the dd value and substitute this value in the equation ax+by=dax+by=d to get the result.
We can write the equation of the line which is parallel to the line 3x+7y=33x+7y=3 as 3x+7y=d3x+7y=d.
Substituting the point (6,7)\left( -6,7 \right) in the equation 3x+7y=d3x+7y=d, then we will get
3(6)+7(7)=d d=4918 d=31 \begin{aligned} & \Rightarrow 3\left( -6 \right)+7\left( 7 \right)=d \\\ & \Rightarrow d=49-18 \\\ & \Rightarrow d=31 \\\ \end{aligned}
Substituting the dd value in 3x+7y=d3x+7y=d, then we will get the equation of the required line as 3x+7y=313x+7y=31.
From both the methods we got the same result.