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Question: How do you write an equation for the line parallel to that contains \(P\left( {2,7} \right)\)?...

How do you write an equation for the line parallel to that contains P(2,7)P\left( {2,7} \right)?

Explanation

Solution

Given is a linear equation in two variables xx and yy . It is given in the standard form of slope intercept form, which is given by y=mx+cy = mx + c, where the value of the slope is mm, and the value of the y-intercept is equal to cc. Parallel lines have the same slope. Using this information we try to solve the problem.

Complete step-by-step solution:
Here the slope of the given linear equation y=8x6y = - 8x - 6 is :
m=8\Rightarrow m = - 8
Any two parallel lines have the same slope, so the line parallel to the given line y=8x6y = - 8x - 6, is going to have the same slope.
Now let us assume the equation of the new line parallel to the given line is given by:
y=mx+c1\Rightarrow y = mx + {c_1}
Where the value of the slope is the same as the given line y=8x6y = - 8x - 6, so the value of the slope is -8.
So the equation of the new parallel line becomes:
y=8x+c1\Rightarrow y = - 8x + {c_1}
Now given that this line passes through the point P(2,7)P\left( {2,7} \right), now substituting this point in the above equation to get the value of c1{c_1}, as shown below:
7=8(2)+c1\Rightarrow 7 = - 8\left( 2 \right) + {c_1}
7=16+c1\Rightarrow 7 = - 16 + {c_1}
On further simplification of the above equation, the value of the new intercept becomes:
c1=23\Rightarrow {c_1} = 23
Now substituting this in the equation y=8x+c1y = - 8x + {c_1}, as shown below:
y=8x+23\Rightarrow y = - 8x + 23
Equation of the line parallel to y=8x6y = - 8x - 6 passing through P(2,7)P\left( {2,7} \right) is y=8x+23y = - 8x + 23.

Note: Please note that we found the parallel line to the given line only when we are given a point passing through it. The most important information is that parallel lines have the same slope, whereas for the perpendicular lines the product of their slopes is equal to -1.