Distance Between a Point and a Line

Notes and Example

\(\text{Step 4: Rewrite in slope intercept form } y=m_2x+b_2\)
\(y-5=\frac{4}{3} (x-4)\)
\(y-5=\frac{4}{3}x -5 \frac{1}{3}\)
\(y=\frac{4}{3}x -\frac{1}{3}\)

\({\text{Distance Between a Point and a Line}}\)
\(\underline{\text{Steps}}\) \(\underline{\text{Example}}\)
\(\text{Given: Point } (x_1,y_1) \text{ and line } y=m_1x+b_1\)
\(\text{Point }(4,5) \text{ and line } y=-\displaystyle\frac{3}{4}x+2\)
\(\text{Step 1: Identify slope of given line}= m_1\)
\(\text{Slope of }y=-\displaystyle\frac{3}{4}x+2 \text{ is }m_1=-\displaystyle\frac{3}{4}\)
\(\text{Step 2: Find perpendicular slope } m_2=-\displaystyle\frac{1}{m_1}\)
\(\text{Perpendicular slope of }m_1=-\displaystyle\frac{3}{4} \text{ is } m_2=\displaystyle\frac{4}{3}\)
\(\text{Step 3: Use point-slope formula with slope } m_2 \text{ and point }(x_1,y_1)\)
\(y-y_1=m_2(x-x_1)\)
\(y-5=\displaystyle\frac{4}{3} (x-4)\)
\(\text{Step 5: Set new equation equal to original equation and solve for }x_2\)
\(m_2 x + b_2 = m_1 x + b_1\)
\(\frac{4}{3}x -\frac{1}{3}=-\frac{3}{4}x+2\)
\(16x -4=-9x+24 \text{ (multiplied both sides by 12)}\)
\(25x -4=24 \)
\(25x =28 \)
\(x_2 =\frac{28}{25} \)
\(\text{Step 6: Plug } x_2 \text{ into either equation to find }y_2\)
\(y_2=-\frac{3}{4}(\frac{28}{25})+2\)
\(y_2=\frac{29}{25}\)
\(\text{Step 7: Find distance between} (x_2,y_2) \text{ and } (x_1,y_1)\)
\(d=\displaystyle\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\(d=\sqrt{(\frac{28}{25}-4)^2+(\frac{29}{25}-5)^2}\)
\(d=\frac{24}{5}=4.8\)
\(\text{Step 8: You are done!}\)
\(\text{The distance between} (4,5) \text{ and } y=-\frac{3}{4}x+2 \text{ is }4.8\)

 

Practice Problems

Find the distance between the given point and line.

\(\textbf{1)}\) \((3,4)\) and \(y=\frac{3}{4}x-2\)Link to Youtube Video Solving Question Number 1

 

\(\textbf{2)}\) \((4,4)\) and \(y=-\frac{4}{3}x-1\)

 

\(\textbf{3)}\) \((0,2)\) and \(y=x\)

 

\(\textbf{4)}\) \((1,-2)\) and \(y=-x-6\)

 

\(\textbf{5)}\) \((-3,4)\) and \(y=-\frac{1}{2}x-2\)

 

 

See Related Pages\(\)

\(\bullet\text{ Geometry Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)
\(\bullet\text{ Graphing Linear Equations}\)
\(\,\,\,\,\,\,\,\,2x-3y=6 \) Thumbnail for Graphing Linear Equations
\(\bullet\text{ Slope Formula}\)
\(\,\,\,\,\,\,\,\,m=\frac{y_2-y_1}{x_2-x_1}\)
\(\bullet\text{ Net Change}\)
\(\,\,\,\,\,\,\,\,y_2-y_1\)
\(\bullet\text{ Slope Intercept Form}\)
\(\,\,\,\,\,\,\,\,y=mx+b\)
\(\bullet\text{ Point Slope Form}\)
\(\,\,\,\,\,\,\,\,y-y_1=m(x-x_1)\)
\(\bullet\text{ Parallel and Perpendicular Slope}\)
\(\,\,\,\,\,\,\,\,m_1=m+2,\,\,\,m_1=\frac{1}{m_2}\)
\(\bullet\text{Finding x- and y- intercepts}\)
\(\,\,\,\,\,\,\,\,y=2x+4\)

 

Wolfram Alpha Calculator

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In Summary

The distance between a point and a line is typically covered in a high school geometry or algebra class.

The distance between a point and a line has many real-world applications. For example, it can be used to measure the proximity of a location to a road or a boundary, or to calculate the shortest distance between two points on a map.

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