Answer:
The equation of required line is: [tex]\mathbf{5x-8y+34=0}[/tex]
Step-by-step explanation:
We need to write equation in the form Ax+By+C=0 of the line parallel to [tex]5x-8y+12=0[/tex] and through the point (-2,3)
First we need to find slope and y-intercept of the required line.
Using equation of line [tex]5x-8y+12=0[/tex] to find slope.
Since the given line and required lines are parallel there slope is same.
Writing equation in slope intercept form: [tex]y=mx+b[/tex] where m is slope.
[tex]5x-8y+12=0 \\-8y=-5x-12\\\frac{-8y}{-8y}=\frac{-5x}{-8}-\frac{12}{-8}\\y=\frac{5}{8}x+\frac{3}{4}[/tex]
So, the slope m = 5/8
The slope of required line is [tex]m=\frac{5}{8}[/tex]
Now finding y-intercept using slope [tex]m=\frac{5}{8}[/tex] and point(-2,3)
[tex]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}[/tex]
So, the equation of required line having slope [tex]m=\frac{5}{8}[/tex] and y-intercept [tex]b=\frac{17}{4}[/tex]
[tex]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[/tex]
So, The equation of required line is: [tex]\mathbf{5x-8y+34=0}[/tex]
