The Notes: `(1, -3), 4x - 3y = -7` Find the distance between the point and the line.

Monday, June 1, 2015

$\displaystyle{\left({1},-{3}\right)},{4}{x}-{3}{y}=-{7}$ Find the distance between the point and the line.

Given (1, -3) $\displaystyle{4}{x}-{3}{y}=-{7}$

$\displaystyle{4}{x}-{3}{y}=-{7}$

$\displaystyle{4}{x}-{3}{y}+{7}={0}$

Let A=4, B=-3, C=7, x1=1, and y1=-3

Find the distance between the point and the line using the formula

$\displaystyle{d}=\frac{{\left|{A}{x}_{{1}}+{B}{y}_{{1}}+{C}\right|}}{\sqrt{{{{A}}^{{2}}+{{B}}^{{2}}}}}$

$\displaystyle{d}=\frac{{\left|{\left({4}\right)}{\left({1}\right)}+{\left(-{3}\right)}{\left(-{3}\right)}+{7}\right|}}{\sqrt{{{{4}}^{{2}}+{{\left(-{3}\right)}}^{{2}}}}}$

$\displaystyle{d}=\frac{{\left|{4}+{9}+{7}\right|}}{\sqrt{{{16}+{9}}}}=\frac{{20}}{{5}}={4}$

The distance between the point and the line is 4 units.

The Notes

Monday, June 1, 2015

$\displaystyle{\left({1},-{3}\right)},{4}{x}-{3}{y}=-{7}$ Find the distance between the point and the line.

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