Answer to Question #134444 in Analytic Geometry for Kimberly

Question

Determine the equation of the line that has a distance 2 from the origin and is
parallel to the line that passes thru P1(2,7) and P2(-3,-2).

Answer 1

calculate the vector

$\overline{P_1P_2} = (P_{2x} - P_{1x};P_{2y} - P_{1y} ) = (-3-2;-2-7) = (-5;-9)$

let ax + by + c =0 is equation of the line

then direction vector is (b;-a) and coincides with the vector $\overline{P_1P_2}$

hence b = -5

-a = -9

a = 9

9x- 5y + c = 0 equation of the line

$h(ax+bx+c=0,(x_0,y_0)) = \frac{|ax_0+by_0+c|} {\sqrt{a^2+b^2}}$

where h is the distance from point to line

in our case h is equal to 2 and the coordinates of the point are equal to zero since they are the origin of coordinates

$\frac{|ax_0+by_0+c|} {\sqrt{a^2+b^2}} = \frac{|c|} {\sqrt{9^2+5^2}} = \frac{|c|} {\sqrt{106}}$

hence

$c = 2\sqrt{106}$ or $c = -2\sqrt{106}$ and equation of the line

$9x -5y + 2\sqrt{106} = 0$

or $9x -5y - 2\sqrt{106} = 0$

Answer: $9x -5y + 2\sqrt{106} = 0$ or $9x -5y - 2\sqrt{106} = 0$