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What is the equation of the line parallel to 3x + 5y = 11 that passes through the point (15, 4)?
O y = -29
= 8x
2 O yx-
y = 3x-21
O x- Ex+5
Oy-+13


Please help!!


Sagot :

Answer:

The equation of the line is [tex]y = -\frac{3x}{5} + 13[/tex]

Step-by-step explanation:

Equation of a line:

The equation of a line has the following format:

[tex]y = mx + b[/tex]

In which m is the slope and b is the y-intercept.

Parallel lines:

If two lines are parallel, their slopes are the same.

Parallel to 3x + 5y = 11

Placing in the standard format:

[tex]5y = -3x + 11[/tex]

[tex]y = -\frac{3x}{5} + \frac{11}{5}[/tex]

So the slope is [tex]-\frac{3}{5}[/tex], which means that the desired equation is:

[tex]y = -\frac{3x}{5} + b[/tex]

Passes through the point (15, 4).

This mean that when [tex]x = 15, y = 4[/tex]. We use this to find b. So

[tex]y = -\frac{3x}{5} + b[/tex]

[tex]4 = -\frac{3(15)}{5} + b[/tex]

[tex]-9 + b = 4[/tex]

[tex]b = 13[/tex]

So

[tex]y = -\frac{3x}{5} + 13[/tex]