Question

3x - 4y + 3 = 0
6x - 8y +7 = 0
a. Find the distance between the two lines.
b. Find the equation of the perpendicular line passing through. (-6,4)
c. Determine the distance of 2 given equations to point (-6.4)

Answer 1

Let two lines be defined by equations $A_1x+B_1y+C_1=0$ and $A_2x+B_2y+C_2=0$ . These lines are parallel if $A_2B_1-A_1B_2=0.$ For our lines $6\cdot(-4)-3\cdot(-8)=0$ , so they are parallel. Let us rewrite the equation of the second line as

$3x-4y+3.5=0.$

Therefore, our lines are defined by equations

$A_1x+B_1y+C_1=0, \;\; A_1x+B_1y+C=0.$

a) Now the distance can be calculated as (see https://en.wikipedia.org/wiki/Distance_between_two_straight_lines)

$d= \dfrac{|C_1-C|}{\sqrt{A^2+B^2}} = \dfrac{|3-3.5|}{\sqrt{3^2+4^2}} = 0.1.$

b) Let the line be defined by equation $A_0x+B_0y+C_0=0.$ This line is perpendicular to $A_1x+B_1y+C_1=0$ if $A_0A_1+B_0B_1 = 0$ . Let us choose $A_0 = 4, \; B_0= 3.$

We know that $A_0\cdot(-6)+B_0\cdot4+C_0=0.$ Therefore, $C_0=12.$

So the equation of line is $4x+3y+12=0.$

c) Distance from point $(x_0,y_0)$ to line $Ax+By+C=0$ (see https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line)

$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$ .

For the first line it is

$d_1=\dfrac{|3\cdot(-6)-4\cdot4+3|}{\sqrt{3^2+4^2}} = \dfrac{31}{5} =6.2 .$

For the second line it is

$d_2=\dfrac{|3\cdot(-6)-4\cdot4+3.5|}{\sqrt{3^2+4^2}} = \dfrac{30.5}{5} = 6.1.$