Shortest Distance from Origin - 2D

The following solutions shows various methods to solve this problem. They all look different, but they essentially use the same concept.

Let the shortest distance of the given line from the origin be $d$ .

Method 1: Distance of line from point formula.

For a line $a x + by + c =0$ , the shortest distance from a point $(x_{0}, y_{0} )$ has the formula

$\text{Shortest Distance} = \frac{|ax_{0} + by_{0} + c | }{\sqrt{ a^{2} + b^{2} }}$

The coordinates of the origin are $(0,0)$ . Our line is $4x - 3y - 12 = 0$ . We can plug these values in the formula to get the distance.

$\begin{aligned} d & = \frac{|4 \times 0 - 3 \times 0 - 12 | }{\sqrt{ 4^{2} + (-3)^{2} }} \\ & = \frac{|0 + 0 - 12 | }{\sqrt{25}} \\ & = \frac{12}{5} = \boxed{2.4} ~~~~ _\square \end{aligned}$

The general form of this formula is proved in this note .

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Method 2: Using area of triangle

If the line does not pass through the origin and if the line is not parallel to either of the coordinate axes, then it will intersect each of the coordinate axes once. Using three points: the origin, the $x$ -intercept and the $y$ -intercept, we can form a right triangle.

The above figure shows the graph of $4x - 3y = 12$ . (The triangle is highlighted in blue). Let the $x$ intercept of the line be $p$ and the $y$ intercept be $q$ . In this problem, $p = 3$ and $q = -4$ .

We can write the area of the triangle in two different ways.

$\text{Area} = \frac{1}{2} \times |p| \times |q| = \frac{1}{2} \times d \times \sqrt{p^{2} + q^{2} }$

From this, we can see that

$\begin{aligned} d &= \frac{|p||q|}{\sqrt{p^{2} + q^{2} }} \\ & = \frac{|3| \times |-4| }{\sqrt{ 3^{2} + (-4)^{2} }} \\ & = \frac{12 }{\sqrt{25}} \\ & = \frac{12}{5} = \boxed{2.4} ~~~~ _\square \end{aligned}$