Question

Line l contains points (0,-2) and (6,6). Point P has the coordinates (-1,5). Find the distance between line l and point p.

Answer 1

d = √  [ (∆x)² + (∆y)²  ]

   d = √  [  (3.2 – 2)² + (6 – 3.6)²   ]

   d = √  7.2  =  √  (36 ⁄ 5)  =  6(√ 5) ⁄ 5  =  2.68 Hope this helps :)

Answer 2

The distance between line l and point p is 5 units.

Given:

Line l contains points $(0,-2)$ and $(6,6)$.

Point P has the coordinates $(-1,5)$.  

To find:

The distance between line l and point P.

Explanation:

The equation of the line l is:

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

$y-(-2)=\dfrac{6-(-2)}{6-0}(x-0)$

$y+2=\dfrac{8}{6}x$

Multiply both sides by $6$.

$6(y+2)=8x$

$6y+12=8x$

$0=8x-6y-12$

The distance between the line $ax+by+c=0$ and the point $(x_0,y_0)$ is:

$D=\dfrac{ax_0+by_0+c}{\sqrt{a^2+b^2}}$

The distance between the line $8x-6y-12=0$ and the point $(-1,5)$.

$D=\dfrac{8(-1)-6(5)-12}{\sqrt{(8)^2+(6)^2}}$

$D=\dfrac{-8-30-12}{\sqrt{64+36}}$

$D=\dfrac{50}{\sqrt{100}}$

$D=\dfrac{50}{10}$

$D=5$

Therefore, the distance between line l and point p is 5 units.

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