Answer to Question #114123 in Analytic Geometry for Randal Rodriguez

Question #114123
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)
1
Expert's answer
2020-05-09T15:29:46-0400

Given Line L1: 3x-4y+3=0 and L2: 6x-8y+7=0     3x4y+(7/2)=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 3(7/2)32+42=0.55=0.1\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)     4x+3y+12=0\implies 4x+3y+12=0 .

c) Distance of line L1 from (-6,4) = 3(6)4(4)+332+42=315.\frac{|3(-6)-4(4)+3|}{\sqrt{3^2+4^2}} = \frac{31}{5}.

Distance of line L2 from(-6,4) = 6(6)8(4)+762+82=6110.\frac{|6(-6)-8(4)+7|}{\sqrt{6^2+8^2}} = \frac{61}{10}.



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