Answer to Question #114123 in Analytic Geometry for Randal Rodriguez

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

Given Line L₁: 3x-4y+3=0 and L₂: 6x-8y+7=0 $\implies 3x-4y+(7/2)=0$.

As clear both lines have the same slope m= 3/4, so given lines are parallell to each other.

a) So, Distance between given lines is $\frac{|3 - (7/2)|}{\sqrt{3^2+4^2}}= \frac{0.5}{5} = 0.1$

b) Now, the slope of perpendicular line to L₁ is -1/m = -4/3 and given perpendicular line is passing through (-6,4).

So, the equation of perpendicular line is (y-4)=(-4/3)(x+6) $\implies 4x+3y+12=0$ .

c) Distance of line L₁ from (-6,4) = $\frac{|3(-6)-4(4)+3|}{\sqrt{3^2+4^2}} = \frac{31}{5}.$

Distance of line L₂ from(-6,4) = $\frac{|6(-6)-8(4)+7|}{\sqrt{6^2+8^2}} = \frac{61}{10}.$