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

Accepted Solution

Answer:The distance between parallel lines is [tex]0.6860\ units[/tex]Step-by-step explanation:we have[tex]3x-5y=1[/tex] -----> equation A[tex]3x-5y=-3[/tex] -----> equation BLine A and Line B are parallel lines, because their slopes are equalStep 1Find the slope of the given linesisolate the variable y in the equation A[tex]5y=3x-1[/tex][tex]y=(3/5)x-1/5[/tex]The slope is [tex]m=(3/5)[/tex]Step 2Find the slope of the perpendicular line to the given lineswe know thatIf two lines are perpendicular, then the product of their slopes is equal to -1so[tex]m1*m2=-1[/tex]we have [tex]m1=3/5[/tex] -----> given linessubstitute[tex](3/5)*m2=-1[/tex] [tex]m2=-5/3[/tex] Step 3Find the equation of the perpendicular line that pass though the origin  with the slope  [tex]m2=-5/3[/tex] and the point (0,0) Find the equation of the line[tex]y-y1=m(x-x1)[/tex][tex]y-0=(-5/3)(x-0)[/tex][tex]y=-(5/3)x[/tex]Step 4Using a graphing toolGraph the two parallel lines and the perpendicular lineThe intersection points are[tex]A(-0.2647,0.4412)[/tex] and [tex]B(0.0882,-0.1471)[/tex]  see the attached figurewe know thatthe distance between the points A and B is the distance between parallel linesthe 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{(-0.1471-0.4412)^{2}+(0.0882+0.2647)^{2}}[/tex][tex]d=\sqrt{(-0.5883)^{2}+(0.3529)^{2}}[/tex][tex]d=0.6860\ units[/tex]