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

THE DISTANCE

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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

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