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Question: How do you find an equation of a line containing the point \(\left( 3,2 \right)\) and parallel to th...

How do you find an equation of a line containing the point (3,2)\left( 3,2 \right) and parallel to the line y2=23xy-2=\dfrac{2}{3}x?

Explanation

Solution

Change of form of the given equation y2=23xy-2=\dfrac{2}{3}x will give the slope of the line. We change it to the form of y=mx+ky=mx+k to find the slope m. Then, we get into the form of its parallel line to find the equation. We put the point value of (3,2)\left( 3,2 \right) to find the value of the constant. We get the equation of the parallel line.

Complete step by step answer:
The given equation of the line y2=23xy-2=\dfrac{2}{3}x can be changed to y=23x+2y=\dfrac{2}{3}x+2.
The given equation y=23x+2y=\dfrac{2}{3}x+2 is of the form y=mx+ky=mx+k. m is the slope of the line.
This gives that the slope of the line y2=23xy-2=\dfrac{2}{3}x is 23\dfrac{2}{3}.
We know that any line parallel to the given line will be of the same slope.
This means any line parallel to y2=23xy-2=\dfrac{2}{3}x will have a slope of 23\dfrac{2}{3}.
We take the equation of any parallel line to y=mx+ky=mx+k as y=mx+cy=mx+c.
Let’s assume that the parallel line to y=23x+2y=\dfrac{2}{3}x+2 as y=23x+py=\dfrac{2}{3}x+p. Here pp is a constant value which we have to find out.
The parallel line goes through the point (3,2)\left( 3,2 \right). We place the point in the equation y=23x+py=\dfrac{2}{3}x+p.
2=23×3+p p=22=0 \begin{aligned} & 2=\dfrac{2}{3}\times 3+p \\\ & \Rightarrow p=2-2=0 \\\ \end{aligned}
Value of pp is 0.
The equation of the line is y=23xy=\dfrac{2}{3}x. Simplified form is 3y=2x3y=2x.

Note: A line parallel to the X-axis does not intersect the X-axis at any finite distance and hence we cannot get any finite x-intercept of such a line. Same goes for lines parallel to the Y-axis. In case of slope of a line the range of the slope is 0 to \infty .