SOLUTION: hi, i am asked to find the standard form of an equationfor a line using the points (9,2) and (5,- 3). can anyone please help me as i am not sure how to do this. thanks

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: hi, i am asked to find the standard form of an equationfor a line using the points (9,2) and (5,- 3). can anyone please help me as i am not sure how to do this. thanks      Log On


   

Question 85093: hi, i am asked to find the standard form of an equationfor a line using the points (9,2) and (5,- 3). can anyone please help me as i am not sure how to do this. thanks
Answer by jim_thompson5910(35256) 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 (9,2) and (5,-3)


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 (9,2) and (x%5B2%5D,y%5B2%5D) is the second point (5,-3))


m=%28-3-2%29%2F%285-9%29 Plug in y%5B2%5D=-3,y%5B1%5D=2,x%5B2%5D=5,x%5B1%5D=9 (these are the coordinates of given points)


m=+-5%2F-4 Subtract the terms in the numerator -3-2 to get -5. Subtract the terms in the denominator 5-9 to get -4




m=5%2F4 Reduce



So the slope is

m=5%2F4





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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-2=%285%2F4%29%28x-9%29 Plug in m=5%2F4, x%5B1%5D=9, and y%5B1%5D=2 (these values are given)



y-2=%285%2F4%29x%2B%285%2F4%29%28-9%29 Distribute 5%2F4


y-2=%285%2F4%29x-45%2F4 Multiply 5%2F4 and -9 to get -45%2F4

y=%285%2F4%29x-45%2F4%2B2 Add 2 to both sides to isolate y


y=%285%2F4%29x-37%2F4 Combine like terms -45%2F4 and 2 to get -37%2F4 (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 (9,2) and (5,-3) is:y=%285%2F4%29x-37%2F4


The equation is now in y=mx%2Bb form (which is slope-intercept form) where the slope is m=5%2F4 and the y-intercept is b=-37%2F4


Notice if we graph the equation y=%285%2F4%29x-37%2F4 and plot the points (9,2) and (5,-3), we get this: (note: if you need help with graphing, check out this solver)


Graph of y=%285%2F4%29x-37%2F4 through the points (9,2) and (5,-3)


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





However, that equation is in slope-intercept form. Lets convert it into standard form

Solved by pluggable solver: Converting Linear Equations in Standard form to Slope-Intercept Form (and vice versa)
Convert from slope-intercept form (y = mx+b) to standard form (Ax+By = C)


y+=+%285%2F4%29x-37%2F4 Start with the given equation


4%2Ay+=+4%2A%28%285%2F4%29x-37%2F4%29 Multiply both sides by the LCD 4


4y+=+5x-37 Distribute and multiply


4y-5x+=+5x-37-5x Subtract 5x from both sides


-5x%2B4y+=+-37 Simplify


-1%2A%28-5x%2B4y%29+=+-1%2A%28-37%29 Multiply both sides by -1 to make the A coefficient positive (note: this step may be optional; it will depend on your teacher and/or textbook)


5x-4y+=+37 Distribute and simplify


The original equation y+=+%285%2F4%29x-37%2F4 (slope-intercept form) is equivalent to 5x-4y+=+37 (standard form where A > 0)


The equation 5x-4y+=+37 is in the form Ax%2BBy+=+C where A+=+5, B+=+-4 and C+=+37