Question 457117
Write an equation of the line containing the given point and perpendicular to the give line. (3, -7); 3x + 2y = 5
:
Find the slope of the given equation, put it in the slope/intercept form: (y = mx+b)
3x + 2y = 5
Subtract 3x from both sides
2y = -3x + 5
divide both by 2
y = {{{-3/2}}}x + {{{5/2}}}
m1 = {{{-3/2}}}
The slope relationship of perpendicular lines: mi*m2 = -1
{{{-3/2}}}*m2 = -1
m2 = -1 * {{{-2/3}}}
m2 = {{{2/3}}} is the slope of the perpendicular line
Find the equation of the perpendicular line using the point/slope form (y-y1=m(x-x1)
x1=3, y1=-7, m = {{{2/3}}}
:
y - (-7) = {{{2/3}}}(x - 3)
y + 7 = {{{2/3}}}x - 3*{{{2/3}}}
y = {{{2/3}}}x - 2 - 7
y = {{{2/3}}}x - 9, the equation perpendicular to 3x + 2y = 5
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Also write an equation of the line containing the given point and parallel to the given line. (-4,3); 6x = 5y+8.
Find the slope the of the given equation
6x = 5y + 8
5y = 6x - 8
Divide both sides by 5
y = {{{6/5}}}x - {{{8/5}}}
parallel lines have the same slope, therefore:
y - 3 = {{{6/5}}}(x - (-4))
y - 3 = {{{6/5}}}(x + 4)
see if you can finish this the same way we did the first problem