Question

Find the equation of the line parallel to the graph of 5x−3y=15 that has an x−intercept of 5.

Answer 1

Equation of the given line is : 5x-3y = 15

i.e., 3y = 5x-15

i.e., y = (5/3)x-5

Then, the slope of the given line is = 5/3.

Now, equation of the line parallel to the given line is : y = (5/3)x+c..........(i)

Since the line (i) has an x-intercept of 5 ,i.e., the line (i) passes through the point (5,0).

Then, 0 = (5/3)×5+c

i.e., 0=(25/3)+c

i.e., c = -25/3

Then the equation (i) becomes, y = (5/3)x+(-25/3)

i.e., 3y = 5x-25

i.e., 5x-3y = 25

Therefore, equation of the required line is : 5x-3y = 25.