Question

Which is the distance between parallel lines with the equations 3x-5y=1 and 3x-5y=-3

Answer 1

Answer:

The distance between parallel lines is $0.6860\ units$

Step-by-step explanation:

we have

$3x-5y=1$ -----> equation A

$3x-5y=-3$ -----> equation B

Line A and Line B are parallel lines, because their slopes are equal

Step 1

Find the slope of the given lines

isolate the variable y in the equation A

$5y=3x-1$

$y=(3/5)x-1/5$

The slope is $m=(3/5)$

Step 2

Find the slope of the perpendicular line to the given lines

we know that

If two lines are perpendicular, then the product of their slopes is equal to -1

so

$m1*m2=-1$

we have

$m1=3/5$ -----> given lines

substitute

$(3/5)*m2=-1$

$m2=-5/3$

Step 3

Find the equation of the perpendicular line that pass though the origin

 with the slope  $m2=-5/3$ and the point (0,0) Find the equation of the line

$y-y1=m(x-x1)$

$y-0=(-5/3)(x-0)$

$y=-(5/3)x$

Step 4

Using a graphing tool

Graph the two parallel lines and the perpendicular line

The intersection points are

$A(-0.2647,0.4412)$ and $B(0.0882,-0.1471)$  

see the attached figure

we know that

the distance between the points A and B is the distance between parallel lines

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$d=\sqrt{(-0.1471-0.4412)^{2}+(0.0882+0.2647)^{2}}$

$d=\sqrt{(-0.5883)^{2}+(0.3529)^{2}}$

$d=0.6860\ units$