Question

What is the equation of a line that passes through the point (8,-2) and is parallel to the line whose equations is 3x+4y=15

Answer 1

For this case we have:

When two lines are parallel, their slopes are equal.

Be a line of the form$y = mx + b$

Where:

m is the slope

b is the cut point

If we have:$3x + 4y = 15$

We can rewrite it as:

$4y = 15-3x$

$y = - \frac {3} {4} x + \frac {15} {4}$

Thus, the slope of that line is given by $m = - \frac {3} {4}$

Since that line is parallel to the one we want to find, then$m = - \frac {3} {4}$is the same for both lines.

The equation of the line that we want to find follows the form:

$y_{2} = m_{2}x_{2} + b_{2}$

Where $m_{2}= - \frac {3} {4}$

So, we have:

$y_{2} = - \frac {3} {4} x_{2} + b_{2}$

We have as data the point $(x_{2}, y_{2}) = (8, -2)$ that passes through the line we want to find. Substituting the points we find the cut point $b_{2}$:

$-2 = - \frac {3} {4} (8) + b_{2}$

$-2 = -6 + b_{2}\\b_{2} = -2 + 6\\b_{2} = 4$

Thus, the equation of the requested line is given by:

$y_{2} = - \frac {3} {4} x_{2} + 4$

Answer:

$y_{2} = - \frac {3} {4} x_{2} + 4$