Question 189656
Step 1) Find the equation of the line that passes through the points B(5,2) and C(-1,4)



First let's find the slope of the line through the points *[Tex \LARGE \left(5,2\right)] and *[Tex \LARGE \left(-1,4\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(5,2\right)] and *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-1,4\right)].



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



{{{m=(4-2)/(-1-5)}}} Plug in {{{y[2]=4}}}, {{{y[1]=2}}}, {{{x[2]=-1}}}, and {{{x[1]=5}}}



{{{m=(2)/(-1-5)}}} Subtract {{{2}}} from {{{4}}} to get {{{2}}}



{{{m=(2)/(-6)}}} Subtract {{{5}}} from {{{-1}}} to get {{{-6}}}



{{{m=-1/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(5,2\right)] and *[Tex \LARGE \left(-1,4\right)] is {{{m=-1/3}}}



Now let's use the point slope formula:



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



{{{y-2=(-1/3)(x-5)}}} Plug in {{{m=-1/3}}}, {{{x[1]=5}}}, and {{{y[1]=2}}}



{{{y-2=(-1/3)x+(-1/3)(-5)}}} Distribute



{{{y-2=(-1/3)x+5/3}}} Multiply



{{{y=(-1/3)x+5/3+2}}} Add 2 to both sides. 



{{{y=(-1/3)x+11/3}}} Combine like terms. 



{{{y=(-1/3)x+11/3}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(5,2\right)] and *[Tex \LARGE \left(-1,4\right)] is {{{y=(-1/3)x+11/3}}}



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Step 2) Find the perpendicular slope



Note: if you draw the line and plot the points, you'll find that the shortest distance is along a perpendicular line from the point to the given line.



Take note that the slope of {{{y=(-1/3)x+11/3}}} is {{{m=-1/3}}}. Change the sign and flip the fraction to get {{{m=3/1}}} or just {{{m=3}}}



So the perpendicular slope is {{{m=3}}}



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Step 3) Find the equation of the line that has a slope of {{{m=3}}} (the perpendicular slope) and goes through the given point A(-2,-2)



If you want to find the equation of line with a given a slope of {{{m=3}}} which goes through the point (-2,-2) 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--2=(3)(x--2)}}} Plug in {{{m=3}}}, {{{x[1]=-2}}}, and {{{y[1]=-2}}} (these values are given)



{{{y+2=(3)(x--2)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(3)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y+2=3x+(3)(2)}}} Distribute {{{3}}}



{{{y+2=3x+6}}} Multiply {{{3}}} and {{{2}}} to get {{{6}}}



{{{y=3x+6-2}}} Subtract 2 from  both sides to isolate y



{{{y=3x+4}}} Combine like terms {{{6}}} and {{{-2}}} to get {{{4}}} 




So the equation of the line that has a slope of {{{m=3}}} goes through (-2,-2) is {{{y=3x+4}}}


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Step 4) Find the point of intersection between the two lines {{{y=(-1/3)x+11/3}}} and {{{y=3x+4}}}



{{{y=3x+4}}} Start with the second equation.



{{{(-1/3)x+11/3=3x+4}}} Plug in {{{y=(-1/3)x+11/3}}}



{{{-x+11=9x+12}}} Multiply EVERY term by the LCD 3 to clear out the fractions.



{{{-x=9x+12-11}}} Subtract {{{11}}} from both sides.



{{{-x-9x=12-11}}} Subtract {{{9x}}} from both sides.



{{{-10x=12-11}}} Combine like terms on the left side.



{{{-10x=1}}} Combine like terms on the right side.



{{{x=(1)/(-10)}}} Divide both sides by {{{-10}}} to isolate {{{x}}}.



{{{x=-1/10}}} Reduce.



{{{y=3x+4}}} Go back to the second equation.



{{{y=3(-1/10)+4}}} Plug in {{{x=-1/10}}}



{{{y=-3/10+4}}} Multiply



{{{y=37/10}}} Combine like terms.




So the lines intersect at the point *[Tex \LARGE \left(-\frac{1}{10}, \frac{37}{10}\right)] 



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Step 5) Find the distance between the two points (-2,-2) and *[Tex \LARGE \left(-\frac{1}{10}, \frac{37}{10}\right)] 




{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} Start with the distance formula.



{{{d=sqrt((-2--1/10)^2+(-2-37/10)^2)}}} Plug in {{{x[1]=-2}}},  {{{x[2]=-1/10}}}, {{{y[1]=-2}}}, and {{{y[2]=37/10}}}.



{{{d=sqrt((-19/10)^2+(-2-37/10)^2)}}} Subtract {{{-1/10}}} from {{{-2}}} to get {{{-19/10}}}.



{{{d=sqrt((-19/10)^2+(-57/10)^2)}}} Subtract {{{37/10}}} from {{{-2}}} to get {{{-57/10}}}.



{{{d=sqrt(361/100+(-57/10)^2)}}} Square {{{-19/10}}} to get {{{361/100}}}.



{{{d=sqrt(361/100+3249/100)}}} Square {{{-57/10}}} to get {{{3249/100}}}.



{{{d=sqrt(361/10)}}} Add {{{361/100}}} to {{{3249/100}}} to get {{{361/10}}}.



{{{d=6.008}}} Approximate the right side with a calculator



So the distance between the two points is about  6.008 units. 



Rounding this value to the nearest hundredth gives us {{{d=6.01}}}



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Answer:



So the distance from the point A(-2,-2) to the line joining B(5,2) and C(-1.4) is approximately 6.01 units




Note: if you want more practice, check out this <a href="http://www.algebra.com/algebra/homework/Coordinate-system/Distance-between-a-point-and-a-line.solver">solver</a>