SOLUTION: Here is the question I'm stuck on: Give an equation of each line described: a) Through the point (2,5) with slope -3/4 b) through the points (3,-1) and (0,6) c) horizon

Algebra ->  Coordinate-system -> SOLUTION: Here is the question I'm stuck on: Give an equation of each line described: a) Through the point (2,5) with slope -3/4 b) through the points (3,-1) and (0,6) c) horizon      Log On


   

Question 978968: Here is the question I'm stuck on:
Give an equation of each line described:
a) Through the point (2,5) with slope -3/4
b) through the points (3,-1) and (0,6)
c) horizontal line through (-2,-7)
d) line through the point (6,5) and parallel to the line 2x + 3y = 7
e) line through the point (4,1) and perpendicular to the line x + 5y = 1

That is all. Thank you so much for taking time out of your day to help me!
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
a) Through the point (2,5) with slope m=-3%2F4

y=mx%2Bb
if slope m=-3%2F4, we have
y=-%283%2F4%29x%2Bb
now use given point (x,y)=(2,5) and find b
5=-%283%2F4%292%2Bb
5=-%283%2F2%29%2Bb
5%2B%283%2F2%29=b
b=10%2F2%2B3%2F2
b=13%2F2
b=6.5
so, your equation is y=-%283%2F4%29x%2B6.5





b) through the points (3,-1) and (0,6)
Solved by pluggable solver: Find the equation of line going through points
hahaWe are trying to find equation of form y=ax+b, where a is slope, and b is intercept, which passes through points (x1, y1) = (3, -1) and (x2, y2) = (0, 6).
Slope a is .
Intercept is found from equation a%2Ax%5B1%5D%2Bb+=+y%5B1%5D, or -2.33333333333333%2A3+%2Bb+=+6. From that,
intercept b is b=y%5B1%5D-a%2Ax%5B1%5D, or b=-1--2.33333333333333%2A3+=+6.

y=(-2.33333333333333)x + (6)

Your graph:





c) horizontal line through (-2,-7)
For a horizontal line equation will be y=a where a is a constant.
For the line to pass through the point (-2,-7) y should be equal to -7.
Your line should be
y=-7


d) line through the point (6,5) and parallel to the line 2x+%2B+3y+=+7
3y+=-2x%2B+7
y+=-%282%2F3%29x%2B+7%2F3

Solved by pluggable solver: Finding the Equation of a Line Parallel or Perpendicular to a Given Line


Since any two parallel lines have the same slope we know the slope of the unknown line is -2%2F3 (its from the slope of y=%28-2%2F3%29%2Ax%2B7%2F3 which is also -2%2F3). Also since the unknown line goes through (6,5), we can find the equation by plugging in this info into the point-slope formula

Point-Slope Formula:

y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope and (x%5B1%5D,y%5B1%5D) is the given point



y-5=%28-2%2F3%29%2A%28x-6%29 Plug in m=-2%2F3, x%5B1%5D=6, and y%5B1%5D=5



y-5=%28-2%2F3%29%2Ax%2B%282%2F3%29%286%29 Distribute -2%2F3



y-5=%28-2%2F3%29%2Ax%2B12%2F3 Multiply



y=%28-2%2F3%29%2Ax%2B12%2F3%2B5Add 5 to both sides to isolate y

y=%28-2%2F3%29%2Ax%2B12%2F3%2B15%2F3 Make into equivalent fractions with equal denominators



y=%28-2%2F3%29%2Ax%2B27%2F3 Combine the fractions



y=%28-2%2F3%29%2Ax%2B9 Reduce any fractions

So the equation of the line that is parallel to y=%28-2%2F3%29%2Ax%2B7%2F3 and goes through (6,5) is y=%28-2%2F3%29%2Ax%2B9


So here are the graphs of the equations y=%28-2%2F3%29%2Ax%2B7%2F3 and y=%28-2%2F3%29%2Ax%2B9



graph of the given equation y=%28-2%2F3%29%2Ax%2B7%2F3 (red) and graph of the line y=%28-2%2F3%29%2Ax%2B9(green) that is parallel to the given graph and goes through (6,5)




e) line through the point (4,1) and perpendicular to the line x+%2B+5y+=+1
5y+=-x%2B+1
y+=-%281%2F5%29x%2B+1%2F5
Solved by pluggable solver: Finding the Equation of a Line Parallel or Perpendicular to a Given Line


Remember, any two perpendicular lines are negative reciprocals of each other. So if you're given the slope of -1%2F5, you can find the perpendicular slope by this formula:

m%5Bp%5D=-1%2Fm where m%5Bp%5D is the perpendicular slope


m%5Bp%5D=-1%2F%28-1%2F5%29 So plug in the given slope to find the perpendicular slope



m%5Bp%5D=%28-1%2F1%29%285%2F-1%29 When you divide fractions, you multiply the first fraction (which is really 1%2F1) by the reciprocal of the second



m%5Bp%5D=5%2F1 Multiply the fractions.


So the perpendicular slope is 5



So now we know the slope of the unknown line is 5 (its the negative reciprocal of -1%2F5 from the line y=%28-1%2F5%29%2Ax%2B1%2F5). Also since the unknown line goes through (4,1), we can find the equation by plugging in this info into the point-slope formula

Point-Slope Formula:

y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope and (x%5B1%5D,y%5B1%5D) is the given point



y-1=5%2A%28x-4%29 Plug in m=5, x%5B1%5D=4, and y%5B1%5D=1



y-1=5%2Ax-%285%29%284%29 Distribute 5



y-1=5%2Ax-20 Multiply



y=5%2Ax-20%2B1Add 1 to both sides to isolate y

y=5%2Ax-19 Combine like terms

So the equation of the line that is perpendicular to y=%28-1%2F5%29%2Ax%2B1%2F5 and goes through (4,1) is y=5%2Ax-19


So here are the graphs of the equations y=%28-1%2F5%29%2Ax%2B1%2F5 and y=5%2Ax-19




graph of the given equation y=%28-1%2F5%29%2Ax%2B1%2F5 (red) and graph of the line y=5%2Ax-19(green) that is perpendicular to the given graph and goes through (4,1)