# Let P(x_1, y_1) be a point and let l be the line with equation ax+ by +c =0. Show the distance d from P->l is given by: d =(ax_1+ by_1 + c)/sqrt( a^2 +b^2)? Find the distance d of the point P(6,7) from the line l with equation 3x +4y =11?

Nov 16, 2016

$d = 7$

#### Explanation:

Let $l \to a x + b y + c = 0$ and ${p}_{1} = \left({x}_{1} , {y}_{1}\right)$ a point not on $l$.

Supposing that $b \ne 0$ and calling ${d}^{2} = {\left(x - {x}_{1}\right)}^{2} + {\left(y - {y}_{1}\right)}^{2}$ after substituting $y = - \frac{a x + c}{b}$ into ${d}^{2}$ we have

${d}^{2} = {\left(x - {x}_{1}\right)}^{2} + {\left(\frac{c + a x}{b} + {y}_{1}\right)}^{2}$. The next step is find the ${d}^{2}$ minimum regarding $x$ so we will find $x$ such that

$\frac{d}{\mathrm{dx}} \left({d}^{2}\right) = 2 \left(x - {x}_{1}\right) - \frac{2 a \left(\frac{c + a x}{b} + {y}_{1}\right)}{b} = 0$. This occours for

$x = \frac{{b}^{2} {x}_{1} - a b {y}_{1} - a c}{{a}^{2} + {b}^{2}}$ Now, substituting this value into ${d}^{2}$ we obtain

${d}^{2} = {\left(c + a {x}_{1} + b {y}_{1}\right)}^{2} / \left({a}^{2} + {b}^{2}\right)$ so

$d = \frac{c + a {x}_{1} + b {y}_{1}}{\sqrt{{a}^{2} + {b}^{2}}}$

Now given

$l \to 3 x + 4 y - 11 = 0$ and ${p}_{1} = \left(6 , 7\right)$ then

$d = \frac{- 11 + 3 \times 6 + 4 \times 7}{\sqrt{{3}^{2} + {4}^{2}}} = 7$