Distance formula: Difference between revisions
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<cmath>\implies ax_1 + by_1 + c + t \cdot \sqrt{a^2+b^2} = 0,</cmath> | <cmath>\implies ax_1 + by_1 + c + t \cdot \sqrt{a^2+b^2} = 0,</cmath> | ||
So; | |||
<cmath> t \cdot \sqrt{a^2+b^2} = -(ax_1+by_1+c)</cmath> | <cmath> t \cdot \sqrt{a^2+b^2} = -(ax_1+by_1+c)</cmath> | ||
<cmath>\implies t = \dfrac{-(ax_1+by_1+c)}{\sqrt{a^2+b^2}}.</cmath> | <cmath>\implies t = \dfrac{-(ax_1+by_1+c)}{\sqrt{a^2+b^2}}.</cmath> | ||
Therefore the perpendicular distance from <imath>(x_1,y_1)</imath> to the line <imath>ax+by+c = 0</imath> is: | Therefore, the perpendicular distance from <imath>(x_1,y_1)</imath> to the line <imath>ax+by+c = 0</imath> is: | ||
<cmath>|t| = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}.</cmath> | <cmath>|t| = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}.</cmath> | ||
Revision as of 23:44, 10 November 2025
The distance formula is a direct application of the Pythagorean Theorem in the setting of a Cartesian coordinate system. In the two-dimensional case, it says that the distance between two points 
and 
is given by 
. In the 
-dimensional case, the distance between 
and 
is 
.
Shortest distance from a point to a line
The distance between the line 
and point 
is
![\[\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}.\]](http://latex-new.aopstest.com/6/7/e/67e32f132391967044607dbe3bd1a0d6a1af3a91.png)
Proof
The equation 
can be written as 
Thus, the perpendicular line through is:
![\[\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{t}{\sqrt{a^2+b^2}},\]](http://latex-new.aopstest.com/c/e/c/cecc79b1a474ef06f682423f36e8ff9fc5fbe45b.png)
where 
is the parameter.
will be the distance from the point along the perpendicular line to 
.
So ![\[x = x_1 + a \cdot \dfrac{t}{\sqrt{a^2+b^2}}\]](http://latex.artofproblemsolving.com/2/c/c/2ccf383b37c445fd12553d8bfaccd183f5906ab8.png)
and ![\[y = y_1 + b \cdot \dfrac{t}{\sqrt{a^2+b^2}}.\]](http://latex-new.aopstest.com/5/2/2/5224444e299ce9a363ac5decd92ea1816435ded5.png)
This meets the given line , where:
![\[a\left(x_1 + a \cdot \dfrac{t}{\sqrt{a^2+b^2}}\right) + b\left(y_1 + b \cdot \dfrac{t}{\sqrt{a^2+b^2}}\right) + c = 0\]](http://latex.artofproblemsolving.com/f/d/6/fd6ba4df459edd33f5fdf86d7825315d84f04f3f.png)
![\[\implies ax_1 + by_1 + c + \dfrac{t(a^2+b^2)}{\sqrt{a^2+b^2}} + c = 0\]](http://latex.artofproblemsolving.com/8/0/7/8077b2738403e402f33929e8c52c36ed245e5fb8.png)
![\[\implies ax_1 + by_1 + c + t \cdot \sqrt{a^2+b^2} = 0,\]](http://latex-new.aopstest.com/7/4/b/74b8865ab174fcca1ebe2f2bb8f77e6ce0df6136.png)
So;
![\[t \cdot \sqrt{a^2+b^2} = -(ax_1+by_1+c)\]](http://latex.artofproblemsolving.com/0/b/5/0b5ce92e2fad16bfb9c26f893a47b2e0bdadfb3f.png)
![\[\implies t = \dfrac{-(ax_1+by_1+c)}{\sqrt{a^2+b^2}}.\]](http://latex-new.aopstest.com/c/f/f/cff5762ec41833ad9f00bd0f5161ac286ac30e98.png)
Therefore, the perpendicular distance from to the line is:
![\[|t| = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}.\]](http://latex-new.aopstest.com/b/c/9/bc9b6cf8ebff4344e51744286feb6e1d34de5e0d.png)
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