Answer to Question #134446 in Analytic Geometry for Kimberly

Question

Find the equation of the line that is parallel to 4x-5y = 2 and whose distance from the
given line is 5.

Answer 1

we know that distance between the parallel lines $Ax+By+C_1=0 \space and \space Ax+By+C_2=0$

is given by

$$d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$$

The rough diagram of the given question is given below

Equation of any line parallel to $4x-5y-2=0$ is of the form $4x-5y+k=0$, where k is constant and its value can be found out by given other geometrical condition.

Now Applying the formula of distance between the parallel lines, we get

On comparing we get $A=4,B=-5 ,C_1=-2 ,and \space C_2=k \ and \space distance (d) = 5 \space units$

$\therefore d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}} \\ or,5=\frac{|-2-k|}{\sqrt{4^2+(-5)^2}}\\or, 5= \pm\frac{(-k-2)}{\sqrt{16+25}}\\Taking \space '-ve' \space sign ,we \space get\\or, 5=\frac{k+2}{\sqrt{41}}\\or, k+2 = 5\sqrt{41}\\or,k= 5\sqrt{41}-2$

$Taking \space '+ve' \space sign ,we \space get\\ 5=\frac{-(k+2)}{\sqrt{41}}\\or, k+2 = -5\sqrt{41}\\or,k= -5\sqrt{41}-2\\or, k = -(5\sqrt{41} +2)$

So the equations of a required line is

$$4x-5y+(5\sqrt{41}-2)=0.\\or\\4x-5y-(5\sqrt{41}+2)=0.$$

ANSWER