THE DISTANCE OF A POINT FROM A LINE

# THE DISTANCE OF A POINT FROM A LINE

C
Asked by 2 years ago
10 million points

Find the distance of the points (a) $(1,3)$, (b) $(-3,4)$, (c) $(4,-2)$ from the line $2x+3y-6=0$.

THE DISTANCE
chegendungu

C
Answered by 2 years ago
10 million points

The distance of $(x_1,y_1)$ from the line $ax-by+c=0$ is

$\bigg|\frac{ax_1+by_1+c}{\sqrt{(a^2+b^2)}}\bigg|$

Therefore the distance of $(1,3)$, $(-3,4)$, $(4,-2)$ from the line $2x+3y-6=0$ are

$\bigg|\frac{2\times 1+3\times3-6}{\sqrt{(2^2+3^2)}}\bigg|=\frac{5}{\sqrt{13}}$

$\bigg|\frac{2\times(-3)+3\times4-6}{\sqrt{(2^2+3^2)}}\bigg|=0$

$\bigg|\frac{2\times 4+3\times(-2)-6}{\sqrt{(2^2+3^2)}}\bigg|=\frac{4}{\sqrt{13}}$

The formula is more easily remembered if two points are noticed:

(1)The numerator is obtained by substituting the coordinates of the point into the equation of the line (remember that the perpendicular distance is zero if the point lies on the line)

(2)The denominator is the square root of the sum of the squares of the coefficients

Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$

Use LaTeX to type formulas and markdown to format text. See example.