Question 173208
# 1


Problem:


What is the equation in slope intercept form of the line that passes through (5,-7) and (2,1)? 




Solution:





First let's find the slope of the line through the points *[Tex \LARGE \left(5,-7\right)] and *[Tex \LARGE \left(2,1\right)]



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



{{{m=(1--7)/(2-5)}}} Plug in {{{y[2]=1}}}, {{{y[1]=-7}}}, {{{x[2]=2}}}, and {{{x[1]=5}}}



{{{m=(8)/(2-5)}}} Subtract {{{-7}}} from {{{1}}} to get {{{8}}}



{{{m=(8)/(-3)}}} Subtract {{{5}}} from {{{2}}} to get {{{-3}}}



{{{m=-8/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(5,-7\right)] and *[Tex \LARGE \left(2,1\right)] is {{{m=-8/3}}}



Now let's use the point slope formula:



{{{y-y[1]=m(x-x[1])}}} Start with the point slope formula



{{{y--7=(-8/3)(x-5)}}} Plug in {{{m=-8/3}}}, {{{x[1]=5}}}, and {{{y[1]=-7}}}



{{{y+7=(-8/3)(x-5)}}} Rewrite {{{y--7}}} as {{{y+7}}}



{{{y+7=(-8/3)x+(-8/3)(-5)}}} Distribute



{{{y+7=(-8/3)x+40/3}}} Multiply



{{{y=(-8/3)x+40/3-7}}} Subtract 7 from both sides. 



{{{y=(-8/3)x+19/3}}} Combine like terms. note: If you need help with fractions, check out this <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">solver</a>.



{{{y=(-8/3)x+19/3}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(5,-7\right)] and *[Tex \LARGE \left(2,1\right)] is {{{y=(-8/3)x+19/3}}}




<hr>



# 2



Problem:



What is the equation in standard form that passes through the point (-2,6) and has a slope of 4





Solution:





If you want to find the equation of line with a given a slope of {{{4}}} which goes through the point (-2,6), you can simply use the point-slope formula to find the equation:



---Point-Slope Formula---
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(x_{1},y_{1}\right)] is the given point


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



{{{y-6=(4)(x--2)}}} Plug in {{{m=4}}}, {{{x[1]=-2}}}, and {{{y[1]=6}}} (these values are given)



{{{y-6=(4)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y-6=4x+(4)(2)}}} Distribute {{{4}}}



{{{y-6=4x+8}}} Multiply {{{4}}} and {{{2}}} to get {{{8}}}



{{{y-6-4x=8}}} Subtract {{{4x}}} from both sides.



{{{y-4x=8+6}}} Add 6 to both sides.



{{{-4x+y=14}}} Combine like terms.



{{{4x-y=-14}}} Multiply EVERY term by -1 to make the "x" coefficient positive.


-----------------------------------


Answer:



So the equation in standard form that has a slope of 4 and passes through the point (-2,6) is 



{{{4x-y=-14}}}