Parallel/Perpendicular Through Point
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What is an equation of the line that passes through the point $(4,0)$ and is perpendicular to the line $4x-3y=15$ ?
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Step 1 ：The equation of a line in the slope-intercept form is given by $y=mx+c$, where $m$ is the slope of the line and $c$ is the y-intercept.

Step 2 ：The slope of a line given by the equation $ax+by=c$ is $-a/b$. Therefore, the slope of the line $4x-3y=15$ is $4/3$.

Step 3 ：The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line. Therefore, the slope of the line we are looking for, which is perpendicular to the line $4x-3y=15$, is $-3/4$.

Step 4 ：We also know that this line passes through the point $(4,0)$. We can substitute these values into the equation $y=mx+c$ to find the y-intercept $c$.

Step 5 ：$m=-0.75$

Step 6 ：$c=3.0$

Step 7 ：Now that we have the slope and the y-intercept, we can write the equation of the line in the slope-intercept form.

Step 8 ：The equation of the line that passes through the point $(4,0)$ and is perpendicular to the line $4x-3y=15$ is $\boxed{{\displaystyle y=-\frac{3}{4}x+3}}$.