Question 98453
These explanations are quite lengthy, so here's a mini table of contents

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<a href=#three>#3</a>



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"#1:
Write the equation of the line passing through the points listed below. Write the final answer in the slope-intercept form y = mx + b. If the slope (m) or the y-intercept (b) are not integers, fractions must be used.
(-1, 8); (9, 2)"




First lets find the slope through the points ({{{-1}}},{{{8}}}) and ({{{9}}},{{{2}}})


{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{-1}}},{{{8}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{9}}},{{{2}}}))


{{{m=(2-8)/(9--1)}}} Plug in {{{y[2]=2}}},{{{y[1]=8}}},{{{x[2]=9}}},{{{x[1]=-1}}}  (these are the coordinates of given points)


{{{m= -6/10}}} Subtract the terms in the numerator {{{2-8}}} to get {{{-6}}}.  Subtract the terms in the denominator {{{9--1}}} to get {{{10}}}

  


{{{m=-3/5}}} Reduce

  

So the slope is

{{{m=-3/5}}}


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




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


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


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



{{{y-8=(-3/5)(x+1)}}} Rewrite {{{x--1}}} as {{{x+1}}}



{{{y-8=(-3/5)x+(-3/5)(1)}}} Distribute {{{-3/5}}}


{{{y-8=(-3/5)x-3/5}}} Multiply {{{-3/5}}} and {{{1}}} to get {{{-3/5}}}


{{{y=(-3/5)x-3/5+8}}} Add {{{8}}} to  both sides to isolate y


{{{y=(-3/5)x+37/5}}} Combine like terms {{{-3/5}}} and {{{8}}} to get {{{37/5}}} (note: if you need help with combining fractions, check out this <a href=http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver>solver</a>)



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Answer:



So the equation of the line which goes through the points ({{{-1}}},{{{8}}}) and ({{{9}}},{{{2}}})  is:{{{y=(-3/5)x+37/5}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=-3/5}}} and the y-intercept is {{{b=37/5}}}


Notice if we graph the equation {{{y=(-3/5)x+37/5}}} and plot the points ({{{-1}}},{{{8}}}) and ({{{9}}},{{{2}}}),  we get this: (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)


{{{drawing(500, 500, -5, 13, -4, 14,
graph(500, 500, -5, 13, -4, 14,(-3/5)x+37/5),
circle(-1,8,0.12),
circle(-1,8,0.12+0.03),
circle(9,2,0.12),
circle(9,2,0.12+0.03)
) }}} Graph of {{{y=(-3/5)x+37/5}}} through the points ({{{-1}}},{{{8}}}) and ({{{9}}},{{{2}}})


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



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#2:


"A line passes through the origin and the point (-6, 4).
Write the equation of the line that passes through (-6, 4) and is perpendicular to the given line.
Write the final answer in the slope-intercept form y = mx + b.
(The slope and y-intercept must be written as fractions, when needed). "



First find the equation of the line through the points (0,0) (the origin) and (-6,4)



First lets find the slope through the points ({{{0}}},{{{0}}}) and ({{{-6}}},{{{4}}})


{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{0}}},{{{0}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{-6}}},{{{4}}}))


{{{m=(4-0)/(-6-0)}}} Plug in {{{y[2]=4}}},{{{y[1]=0}}},{{{x[2]=-6}}},{{{x[1]=0}}}  (these are the coordinates of given points)


{{{m= 4/-6}}} Subtract the terms in the numerator {{{4-0}}} to get {{{4}}}.  Subtract the terms in the denominator {{{-6-0}}} to get {{{-6}}}

  


{{{m=-2/3}}} Reduce

  

So the slope is

{{{m=-2/3}}}


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



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




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


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


{{{y-0=(-2/3)(x-0)}}} Plug in {{{m=-2/3}}}, {{{x[1]=0}}}, and {{{y[1]=0}}} (these values are given)



{{{y-0=(-2/3)x+(-2/3)(0)}}} Distribute {{{-2/3}}}


{{{y-0=(-2/3)x+0}}} Multiply {{{-2/3}}} and {{{0}}} to get {{{0/3}}}. Now reduce {{{0/3}}} to get {{{0}}}


{{{y=(-2/3)x+0+0}}} Add {{{0}}} to  both sides to isolate y


{{{y=(-2/3)x+0}}} Combine like terms {{{0}}} and {{{0}}} to get {{{0}}} 

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Answer:



So the equation of the line which goes through the points ({{{0}}},{{{0}}}) and ({{{-6}}},{{{4}}})  is:{{{y=(-2/3)x}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=-2/3}}} and the y-intercept is {{{b=0}}}


Notice if we graph the equation {{{y=(-2/3)x}}} and plot the points ({{{0}}},{{{0}}}) and ({{{-6}}},{{{4}}}),  we get this: (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)


{{{drawing(500, 500, -12, 6, -7, 11,
graph(500, 500, -12, 6, -7, 11,(-2/3)x+0),
circle(0,0,0.12),
circle(0,0,0.12+0.03),
circle(-6,4,0.12),
circle(-6,4,0.12+0.03)
) }}} Graph of {{{y=(-2/3)x}}} through the points ({{{0}}},{{{0}}}) and ({{{-6}}},{{{4}}})


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


*[invoke equation_parallel_or_perpendicular "perpendicular", "-2/3", 0, -6, 4]




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#3



"Write the equation for the line perpendicular to
2x - 5y = 5 and passes through the point (-2, -5).
Enter the variable expression Ax + By separately from the constant term C.
Your response should be given in standard form where A, B and C are the smallest possible integers and A > 0.


First convert 2x - 5y = 5 to slope intercept form



*[invoke converting_linear_equations "standard_to_slope-intercept", 2, -5, 5, 2, 1]



*[invoke equation_parallel_or_perpendicular "perpendicular", "2/5", -1, -2, -5]



Now convert {{{(-5/2)x-10}}} back to standard form



*[invoke converting_linear_equations "slope-intercept_to_standard", "-5/2", -5, 5, "-5/2", -10]





If you have any further questions, feel free to ask.