What is the equation of the line parallel to 3x + 5y = 11 that passes through the point (15, 4)

3x + 5y = 11
5y = -3x + 11
y = -3/5x + 11/5.....slope here is -3/5. A parallel line will have the same slope.

y = mx + b
slope(m) = -3/5
(15,4)...x = 15 and y = 4
now we sub and find b, the y int
4 = 15(-3/5) + b
4 = - 9 + b
4 + 9 = b
13 = b

ur equation is : y = -3/5x + 13

Answer Link

ogorwyne ogorwyne · Answer 2 · 2022-07-07T12:15:38+02:00

The equation of the straight line can be solved as y = -3/5x + 13

We have the equation of a straight lien to be give as

y = mx + b

This equation is given as 3x + 5y = 11

We have to make y the subject of the formula

y = -(3x + 11)/5

The slope of the equation is -3x/5 the intercept is 11/5

With x = 15 andd y = 4

We have to put this in the equation

4 = 15(-3/5) + b

4 = -45/5 + b

b = 13

Hence the equation is said to be

y = -3/5x + 13

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