SOLUTION: find the slope of the line that passes through (-3,1) and (2,-6). find an equation of each line in standard form satisfying the given conditions.

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Question 106865This question is from textbook intermediate algebra
: find the slope of the line that passes through (-3,1) and (2,-6).
find an equation of each line in standard form satisfying the given conditions.This question is from textbook intermediate algebra

Found 2 solutions by jim_thompson5910, MathLover1:
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Finding the Equation of a Line
First lets find the slope through the points (-3,1) and (2,-6)


m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29 Start with the slope formula (note: (x%5B1%5D,y%5B1%5D) is the first point (-3,1) and (x%5B2%5D,y%5B2%5D) is the second point (2,-6))


m=%28-6-1%29%2F%282--3%29 Plug in y%5B2%5D=-6,y%5B1%5D=1,x%5B2%5D=2,x%5B1%5D=-3 (these are the coordinates of given points)


m=+-7%2F5 Subtract the terms in the numerator -6-1 to get -7. Subtract the terms in the denominator 2--3 to get 5



So the slope is

m=-7%2F5


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Now let's use the point-slope formula to find the equation of the line:




------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 one of the given points


So lets use the Point-Slope Formula to find the equation of the line


y-1=%28-7%2F5%29%28x--3%29 Plug in m=-7%2F5, x%5B1%5D=-3, and y%5B1%5D=1 (these values are given)



y-1=%28-7%2F5%29%28x%2B3%29 Rewrite x--3 as x%2B3



y-1=%28-7%2F5%29x%2B%28-7%2F5%29%283%29 Distribute -7%2F5


y-1=%28-7%2F5%29x-21%2F5 Multiply -7%2F5 and 3 to get -21%2F5

y=%28-7%2F5%29x-21%2F5%2B1 Add 1 to both sides to isolate y


y=%28-7%2F5%29x-16%2F5 Combine like terms -21%2F5 and 1 to get -16%2F5 (note: if you need help with combining fractions, check out this solver)



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Answer:



So the equation of the line which goes through the points (-3,1) and (2,-6) is:y=%28-7%2F5%29x-16%2F5


The equation is now in y=mx%2Bb form (which is slope-intercept form) where the slope is m=-7%2F5 and the y-intercept is b=-16%2F5


Notice if we graph the equation y=%28-7%2F5%29x-16%2F5 and plot the points (-3,1) and (2,-6), we get this: (note: if you need help with graphing, check out this solver)


drawing%28500%2C+500%2C+-9.5%2C+8.5%2C+-11.5%2C+6.5%2C%0D%0A++graph%28500%2C+500%2C+-9.5%2C+8.5%2C+-11.5%2C+6.5%2C%28-7%2F5%29x%2B-16%2F5%29%2C%0D%0A++circle%28-3%2C1%2C0.12%29%2C%0D%0A++circle%28-3%2C1%2C0.12%2B0.03%29%2C%0D%0A++circle%282%2C-6%2C0.12%29%2C%0D%0A++circle%282%2C-6%2C0.12%2B0.03%29%0D%0A++%29+ Graph of y=%28-7%2F5%29x-16%2F5 through the points (-3,1) and (2,-6)


Notice how the two points lie on the line. This graphically verifies our answer.





Now let's convert the slope-intercept equation into standard form


Solved by pluggable solver: Converting Linear Equations in Standard form to Slope-Intercept Form (and vice versa)


y=%28-7%2F5%29%2Ax-16%2F5Start with the given equation



y%2B%287%2F5%29%2Ax=-16%2F5Add %287%2F5%29%2Ax to both sides

%2B%287%2F5%29%2Ax%2By=-16%2F5 Rearrange

5%28%2B%287%2F5%29%2Ax%2By%29=5%28-16%2F5%29 Multiply both sides by 5



5%28%2B7%2F5%29x%2B5%2Ay=5%28-16%2F5%29 Distribute



%2B%2835%2F5%29%2Ax%2B5%2Ay=-80%2F5 Multiply

%2B7%2Ax%2B5%2Ay=-16 Reduce

Now the equation is in standard form Ax%2BBy=C where A=%2B7, B=5, and C=-16


Answer by MathLover1(3376) About Me  (Show Source):
You can put this solution on YOUR website!
find the slope of the line that passes through (-3,1) and (2,-6).
slope+=+%28change_in_y%2Fchange_in_x%29
If slope+=+m, change_in_y+=+y%5B2%5D+%96+y%5B1%5D, change_in_x+=+x%5B2%5D+%96+x%5B1%5D, then we have:
m =(y[2] – y[1])/(x[2] – x[1])
Since
x%5B1%5D+=-3
x%5B2%5D+=2
y%5B1%5D+=1
y%5B2%5D+=-6

we will have:
m = (-6– 1)/(2 – (-3))
m+=+-7%2F%282+%2B+3%29

m+=+-%287%2F5%29
We are trying to find equation y+=+ax+%2B+b.
The value of slope a+=+-7%2F5 is already given to us, as a point (-3,1) that lies on the line as well.
we need b which is:
b+=+y%5B1%5D – %28ax%5B1%5D%29
b+=+1+-+%28-7%2F5%28-3%29%29
b+=+1+-%2821%2F5%29
b+=+%285-21%29%2F5
b=+-%2816%2F5%29
so,
y+=+ax+%2B+b will be:
y+=+-%287%2F5%29x+-+16%2F5
here is the graph of this function, make sure that both given points (-3,1) and (2,-6)lie on line.

Solved by pluggable solver: Graphing Linear Equations
In order to graph y=-1.4%2Ax-3.2 we only need to plug in two points to draw the line

So lets plug in some points

Plug in x=-8

y=-1.4%2A%28-8%29-3.2

y=11.2-3.2 Multiply

y=8 Add

So here's one point (-8,8)


drawing%28+600%2C+600%2C+-18%2C12%2C+-16%2C+18%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-8%2C8%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-8%2C8%2C+.1%2C+1.5+%29+%29%0D%0A++%29

Now lets find another point

Plug in x=2

y=-1.4%2A%282%29-3.2

y=-2.8-3.2 Multiply

y=-6 Add

So here's another point (2,-6). Add this to our graph


drawing%28+600%2C+600%2C+-18%2C12%2C+-16%2C+18%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+2%2C-6%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+2%2C-6%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-8%2C8%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-8%2C8%2C+.1%2C+1.5+%29+%29%0D%0A++%29


Now draw a line through these points

drawing%28+600%2C+600%2C+-18%2C12%2C+-16%2C+18%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-18-10%2C-1.4%2A%28-18-10%29%2B-3.2%2C12%2B10%2C-1.4%2A%2812%2B10%29%2B-3.2%29%2C%0D%0A++++blue%28+circle%28+2%2C-6%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+2%2C-6%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-8%2C8%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-8%2C8%2C+.1%2C+1.5+%29+%29%0D%0A++%29 So this is the graph of y=-1.4%2Ax-3.2 through the points (-8,8) and (2,-6)


So from the graph we can see that the slope is -1.4%2F1 (which tells us that in order to go from point to point we have to start at one point and go down -1.4 units and to the right 1 units to get to the next point), the y-intercept is (0,-3.2)and the x-intercept is (-2.28571428571429,0)


We could graph this equation another way. Since b=-3.2 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,-3.2).


So we have one point (0,-3.2)


drawing%28+600%2C+600%2C+-10%2C11%2C+-14.6%2C+6.8%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++%0D%0A++++blue%28+circle%28+0%2C-3.2%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+0%2C-3.2%2C+0.14-0.05%29+%29%0D%0A++%29


Now since the slope is -1.4%2F1, this means that in order to go from point to point we can use the slope to do so. So starting at (0,-3.2), we can go down 1.4 units


drawing%28++600%2C+600%2C+-10%2C11%2C+-14.6%2C+6.8%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+0%2C-3.2%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+0%2C-3.2%2C+0.14-0.05%29+%29%0D%0A++blue%28+arc%28+0%2C+-3.2%2B%28-1.4%2F2%29%2C+2%2C+-1.4%2C+90%2C+270+%29+%29%0D%0A++%0D%0A++%29
and to the right 1 units to get to our next point

drawing%28++600%2C+600%2C+-10%2C11%2C+-14.6%2C+6.8%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+1%2C-4.6%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+1%2C-4.6%2C+0.14-0.05+%29+%29%2C%0D%0A++++blue%28+circle%28+0%2C-3.2%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+0%2C-3.2%2C+0.14-0.05%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+-3.2%2B%28-1.4%2F2%29%2C+2%2C+-1.4%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%281%2F2%29%2C+-4.6%2C+1%2C+2%2C+0%2C+180+%29+%29%0D%0A++%0D%0A++%29
Now draw a line through those points to graph y=-1.4%2Ax-3.2


drawing%28+600%2C+600%2C+-10%2C11%2C+-14.6%2C+6.8%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-10-10%2C-1.4%2A%28-10-10%29%2B-3.2%2C11%2B10%2C-1.4%2A%2811%2B10%29%2B-3.2%29%2C%0D%0A++++blue%28+circle%28+1%2C-4.6%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+1%2C-4.6%2C+0.14-0.05+%29+%29%2C%0D%0A++++blue%28+circle%28+0%2C-3.2%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+0%2C-3.2%2C+0.14-0.05%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+-3.2%2B%28-1.4%2F2%29%2C+2%2C+-1.4%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%281%2F2%29%2C+-4.6%2C+1%2C+2%2C+0%2C+180+%29+%29%0D%0A++%29 So this is the graph of y=-1.4%2Ax-3.2 through the points (0,-3.2) and (1,-4.6)