Answer :
Answer:
The distance between the two parallel lines is 11/17 units
Step-by-step explanation:
we have
8x-15y+5=0 -----> equation A
16x-30y-12=0 ----> equation B
Divide by 2 both sides equation B
8x-15y-6=0 ----> equation C
Compare equation A and equation C
Line A and Line C are parallel lines with different y-intercept
step 1
Find the slope of the parallel lines (The slope of two parallel lines is the same)
8x-15y+5=0
15y=8x+5
[tex]y=\frac{8}{15}x+\frac{1}{3}[/tex]
the slope is
[tex]m=\frac{8}{15}[/tex]
step 2
Find the slope of a line perpendicular to the given lines
Remember that
If two lines are perpendicular then their slopes are opposite reciprocal (the product of their slopes is -1)
[tex]m1*m2=-1[/tex]
we have
[tex]m1=\frac{8}{15}[/tex]
therefore
[tex]m2=-\frac{15}{8}[/tex]
step 3
Find the equation of the line perpendicular to the given lines
assume any point that lie on line A
[tex]y=\frac{8}{15}x+\frac{1}{3}[/tex]
For [tex]x=0[/tex]
[tex]y=\frac{1}{3}[/tex]
To find the equation of the line we have
[tex]point\ (0,1/3)[/tex] ---> is the y-intercept
[tex]m=-\frac{15}{8}[/tex]
The equation in slope intercept form is
[tex]y=-\frac{15}{8}x+\frac{1}{3}[/tex] -----> equation D
step 4
Find the intersection point of the perpendicular line with the Line C
we have the system of equations
[tex]y=-\frac{15}{8}x+\frac{1}{3}[/tex] ----> equation D
[tex]8x-15y-6=0[/tex] ----> [tex]y=\frac{8}{15}x-\frac{2}{5}[/tex] ----> equation E
equate equation D and equation E and solve for x
[tex]\frac{8}{15}x-\frac{2}{5}=-\frac{15}{8}x+\frac{1}{3}[/tex]
[tex]\frac{8}{15}x+\frac{15}{8}x=\frac{1}{3}+\frac{2}{5}[/tex]
Multiply by 120 both sides to remove fractions
[tex]64x+225x=40+48[/tex]
[tex]289x=88[/tex]
[tex]x=88/289[/tex]
Find the value of y
[tex]y=-\frac{15}{8}(88/289)+\frac{1}{3}[/tex]
[tex]y=-\frac{206}{867}[/tex]
the intersection point is [tex](\frac{88}{289},-\frac{206}{867})[/tex]
step 5
Find the distance between the points [tex](0,\frac{1}{3})[/tex] and [tex](\frac{88}{289},-\frac{206}{867})[/tex]
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
substitute the values
[tex]d=\sqrt{(-\frac{206}{867}-\frac{1}{3})^{2}+(\frac{88}{289}-0)^{2}}[/tex]
[tex]d=\sqrt{(-\frac{495}{867})^{2}+(\frac{88}{289})^{2}}[/tex]
[tex]d=\sqrt{(\frac{245,025}{751,689})+(\frac{7,744}{83,521})}[/tex]
[tex]d=\sqrt{\frac{314,721}{751,689}}[/tex]
[tex]d=\frac{561}{867}\ units[/tex]
Simplify
[tex]d=\frac{11}{17}\ units[/tex]
therefore
The distance between the two parallel lines is 11/17 units
see the attached figure to better understand the problem
