Question

Please help FAST I need the answer along with you explaining the steps so I can get it thanks! I’ll give Brainlist also!

Answer 1

Answer:

The equation of required line is: $\mathbf{5x-8y+34=0}$

Step-by-step explanation:

We need to write equation in the form Ax+By+C=0 of the line parallel to $5x-8y+12=0$ and through the point (-2,3)

First we need to find slope and y-intercept of the required line.

Using equation of line  $5x-8y+12=0$ to find slope.

Since the given line and required lines are parallel there slope is same.

Writing equation in slope intercept form: $y=mx+b$ where m is slope.

$5x-8y+12=0 \\-8y=-5x-12\\\frac{-8y}{-8y}=\frac{-5x}{-8}-\frac{12}{-8}\\y=\frac{5}{8}x+\frac{3}{4}$

So, the slope m = 5/8

The slope of required line is $m=\frac{5}{8}$

Now finding y-intercept using slope $m=\frac{5}{8}$ and point(-2,3)

$y=mx+b\\3=\frac{5}{8}(-2)+b\\3=\frac{-5}{4}+b\\b=3+\frac{5}{4}\\b=\frac{3*4+5}{4}\\b=\frac{12+5}{4}\\b=\frac{17}{4}$

So, the equation of required line having slope $m=\frac{5}{8}$  and y-intercept $b=\frac{17}{4}$

$y=mx+b\\y=\frac{5}{8}x+\frac{17}{4}\\y=\frac{5x+17*2}{8}\\y= \frac{5x+34}{8} \\8y=5x+34\\5x-8y+34=0$

So, The equation of required line is: $\mathbf{5x-8y+34=0}$