Answer:
Part A) [tex]y=-\frac{A}{B}x+\frac{7B+2A}{B}[/tex]
Part B) [tex]y=\frac{B}{A}x+\frac{7A-2B}{A}[/tex]
Step-by-step explanation:
we have the equation of the line L in standard form
[tex]Ax+By=C[/tex]
isolate the variable y
[tex]By=-Ax+C[/tex]
[tex]y=-(A/B)x+C/B[/tex]
The slope of the line L is equal to
[tex]m=-A/B[/tex]
Part A) Write the equation, in slope intercept form, of the line passing through the point (2, 7) and parallel to line L
Remember that
If two line are parallel, then their slopes are equal
The equation of the line in slope intercept form is equal to
[tex]y=mx+b[/tex]
we have
[tex]m=-A/B[/tex]
[tex]point (2,7)[/tex]
substitute the values and solve for b
[tex]7=(-A/B)(2)+b[/tex]
[tex]b=7+2A/B=(7B+2A)/B[/tex]
The equation of the line is
[tex]y=-\frac{A}{B}x+\frac{7B+2A}{B}[/tex]
Part B) Write the equation, in slope intercept form, of the line passing through the point (2, 7) and perpendicular to line L
Remember that
If two line are perpendicular, then the product of their slopes is equal to -1
[tex]m1*m2=-1[/tex]
The equation of the line in slope intercept form is equal to
[tex]y=mx+b[/tex]
we have
[tex]m1=-A/B[/tex]
Find the value of m2
[tex](-A/B)*m2=-1[/tex]
[tex]m2=B/A[/tex]
[tex]point (2,7)[/tex]
substitute the values and solve for b
[tex]7=(B/A)(2)+b[/tex]
[tex]b=7-2B/A=(7A-2B)/A[/tex]
The equation of the line is
[tex]y=\frac{B}{A}x+\frac{7A-2B}{A}[/tex]
