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Distance of A Point From A Line: Definition and Examples | EDU.COM

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Distance of A Point From A Line: Definition and Examples | EDU.COMEDU.COMResourcesBlogGuidePodcastPlanBackHomesvg]:size-3.5">Math Glossarysvg]:size-3.5">Distance of A Point From A LineDistance of A Point From A Line: Definition and ExamplesTable of ContentsDistance Between Point and Line Definition of Distance Between Point and Line

The distance between a point and a line is defined as the shortest distance from the point to any point on the line. To find this shortest distance, we need to draw a perpendicular from the point to the line. The length of this perpendicular line segment represents the actual distance between the point and the line. When working with a line in the form Ax+By+C=0Ax + By + C = 0Ax+By+C=0 and a point with coordinates (x0,y0)(x_0, y_0)(x0​,y0​), the distance can be calculated using the formula d=∣Ax0+By0+C∣A2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}d=A2+B2​∣Ax0​+By0​+C∣​.

This formula contains an absolute value sign in the numerator, ensuring the distance is always positive or zero. The denominator represents the square root of the sum of squares of the coefficients AAA and BBB. The perpendicular distance calculation is grounded in coordinate geometry principles, involving the relationship between the area of a triangle and the height from a point to a line. The distance formula can also be simplified for special cases, such as finding the distance from the origin to a line, which becomes d=∣C∣A2+B2d = \frac{|C|}{\sqrt{A^2 + B^2}}d=A2+B2​∣C∣​.

Examples of Distance Between Point and Line Example 1: Finding Distance from Origin to a Line Problem:

Find the distance between the point (0,0)(0, 0)(0,0) and the line 4x−3y+25=04x - 3y + 25 = 04x−3y+25=0.

Step-by-step solution:

Step 1, Identify the point and the line equation. We have the point (0,0)(0, 0)(0,0) and the line 4x−3y+25=04x - 3y + 25 = 04x−3y+25=0.

Step 2, Identify the values from the line equation. Compare with the standard form Ax+By+C=0Ax + By + C = 0Ax+By+C=0:

A=4,B=−3,C=25A = 4, B = -3, C = 25A=4,B=−3,C=25

Step 3, Substitute the values into the distance formula:

d=∣(Ax1+By1+C)∣A2+B2d = \frac{|(Ax_1 + By_1 + C)|}{\sqrt{A^2 + B^2}}d=A2+B2​∣(Ax1​+By1​+C)∣​

Step 4, Replace the variables with their values:

d=∣(4×0)+(−3×0)+25∣42+(−3)2d = \frac{|(4 \times 0) + (-3 \times 0) + 25|}{\sqrt{4^2 + (-3)^2}}d=42+(−3)2​∣(4×0)+(−3×0)+25∣​

Step 5, Simplify the expression:

d=∣0+0+25∣16+9=2525=255=5d = \frac{|0 + 0 + 25|}{\sqrt{16 + 9}} = \frac{25}{\sqrt{25}} = \frac{25}{5} = 5d=16+9​∣0+0+25∣​=25​25​=525​=5 units Example 2: Finding Distance from a Point to a Slope-Intercept Form Line Problem:

Find the distance between the point (4,6)(4, 6)(4,6) and line y=13x−3y = \frac{1}{3}x - 3y=31​x−3.

Step-by-step solution:

Step 1, Identify the point coordinates. We have (4,6)(4, 6)(4,6) where x1=4x_1 = 4x1​=4 and y1=6y_1 = 6y1​=6.

Step 2, Convert the line equation to standard form Ax+By+C=0Ax + By + C = 0Ax+By+C=0:

y=13x−3y = \frac{1}{3}x - 3y=31​x−3 Add 3 to both sides: y+3=13xy + 3 = \frac{1}{3}xy+3=31​x Rearrange to get: 13x−y−3=0\frac{1}{3}x - y - 3 = 031​x−y−3=0 Multiply all terms by 3: x−3y−9=0x - 3y - 9 = 0x−3y−9=0

Step 3, Identify the coefficients:

A=1,B=−3,C=−9A = 1, B = -3, C = -9A=1,B=−3,C=−9

Step 4, Substitute the values into the distance formula:

d=∣(Ax1+By1+C)∣A2+B2d = \frac{|(Ax_1 + By_1 + C)|}{\sqrt{A^2 + B^2}}d=A2+B2​∣(Ax1​+By1​+C)∣​ d=∣(1×4)+(−3×6)−9∣(1)2+(−3)2d = \frac{|(1 \times 4) + (-3 \times 6) - 9|}{\sqrt{(1)^2 + (-3)^2}}d=(1)2+(−3)2​∣(1×4)+(−3×6)−9∣​

Step 5, Simplify the expression:

d=∣4−18−9∣1+9=∣−23∣10=2310d = \frac{|4 - 18 - 9|}{\sqrt{1 + 9}} = \frac{|-23|}{\sqrt{10}} = \frac{23}{\sqrt{10}}d=1+9​∣4−18−9∣​=10​∣−23∣​=10​23​ units Example 3: Calculating Distance with Integer Coefficients Problem:

Find the distance between point P(6,9)P(6, 9)P(6,9) and line 2x−5y+7=02x - 5y + 7 = 02x−5y+7=0.

Step-by-step solution:

Step 1, Identify the point coordinates. We have P(6,9)P(6, 9)P(6,9) where x1=6x_1 = 6x1​=6 and y1=9y_1 = 9y1​=9.

Step 2, Identify the values from the line equation. Compare with the standard form Ax+By+C=0Ax + By + C = 0Ax+By+C=0:

A=2,B=−5,C=7A = 2, B = -5, C = 7A=2,B=−5,C=7

Step 3, Substitute the values into the distance formula:

d=∣(Ax1+By1+C)∣A2+B2d = \frac{|(Ax_1 + By_1 + C)|}{\sqrt{A^2 + B^2}}d=A2+B2​∣(Ax1​+By1​+C)∣​ d=∣(2×6)+(−5×9)+7∣(2)2+(−5)2d = \frac{|(2 \times 6) + (-5 \times 9) + 7|}{\sqrt{(2)^2 + (-5)^2}}d=(2)2+(−5)2​∣(2×6)+(−5×9)+7∣​

Step 4, Simplify the expression:

d=∣12−45+7∣4+25=∣−26∣29=2629d = \frac{|12 - 45 + 7|}{\sqrt{4 + 25}} = \frac{|-26|}{\sqrt{29}} = \frac{26}{\sqrt{29}}d=4+25​∣12−45+7∣​=29​∣−26∣​=29​26​ units Comments(4)GGamerZackNovember 6, 2025I've been struggling to explain this to my students. This clear def and examples made it a breeze. Thanks for the great resource!

AAthleteIvyNovember 6, 2025I've been struggling to explain this to my students. This glossary page's clear def and examples made it so much easier! Thanks!

MManagerPaulineNovember 4, 2025This glossary page is great! I've used the distance formula here to help my students grasp the concept. Clear and useful for learning.

NNatureLover75September 17, 2025This explanation really helped my son understand the concept! The formula breakdown was clear, and the examples made it easy to follow. We even used it for his homework, and he got it right!

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