Find the distance from point (1,-\frac{1}{2}) to the line y=5.

Question:

Find the distance from point {eq}\left(1,-\frac{1}{2}\right) {/eq} to the line {eq}y=5 {/eq}.

Distance Between a Point and a Horizontal Line:

The distance between the point {eq}(x_0, y_0) {/eq} and the horizontal line {eq}y = c {/eq} is merely {eq}d= |y_0-c| {/eq}.

To determine the distance between a point and a vertical line {eq}x=b {/eq}, we take the absolute value of the difference between the abscissa of the point and {eq}b {/eq}.

Answer and Explanation:

The point we're given is {eq}\displaystyle \left(1,-\frac{1}{2}\right) {/eq} and its ordinate is {eq}\displaystyle -\frac{1}{2} {/eq} so {eq}y_0= - \displaystyle \frac{1}{2} {/eq} for the formula we'll use to find the distance from {eq}\displaystyle \left(1,-\frac{1}{2}\right) {/eq} to {eq}y=5 {/eq}.

As {eq}y=5 {/eq}, we have {eq}c=5 {/eq} and by plugging in these values we obtain:

{eq}\begin{align*} \displaystyle d&= |y_0-c|\\ d&= \left|- \frac{1}{2}-5\right|\\ d&= \left|- \frac{11}{2}\right|\\ d&=\frac{11}{2}\\ \end{align*} {/eq}

Thus, the distance from the point {eq}\displaystyle \left(1,-\frac{1}{2}\right) {/eq} to the line {eq}y=5 {/eq} is {eq}\displaystyle \frac{11}{2} {/eq} units.


Learn more about this topic:

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How to Find the Distance Between a Point & a Line

from FTCE Mathematics 6-12 (026): Practice & Study Guide

Chapter 30 / Lesson 5
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