Question 171849
Note: I don't mean to be nitpicky, but please use spaces between your problems (as it's hard to find where one problem ends and another begins).


# 1



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



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


{{{y-2=3x-9}}} Multiply {{{3}}} and {{{-3}}} to get {{{-9}}}


{{{y=3x-9+2}}} Add 2 to  both sides to isolate y


{{{y=3x-7}}} Combine like terms {{{-9}}} and {{{2}}} to get {{{-7}}} 

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Answer:



So the equation of the line with a slope of {{{3}}} which goes through the point ({{{3}}},{{{2}}}) is:


{{{y=3x-7}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=3}}} and the y-intercept is {{{b=-7}}}


Notice if we graph the equation {{{y=3x-7}}} and plot the point ({{{3}}},{{{2}}}),  we get (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, -6, 12, -7, 11,
graph(500, 500, -6, 12, -7, 11,(3)x+-7),
circle(3,2,0.12),
circle(3,2,0.12+0.03)
) }}} Graph of {{{y=3x-7}}} through the point ({{{3}}},{{{2}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{3}}} and goes through the point ({{{3}}},{{{2}}}), this verifies our answer.



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# 2



{{{2x+y=4}}} Start with the given equation.



{{{y=4-2x}}} Subtract {{{2x}}} from both sides.



{{{y=-2x+4}}} Rearrange the terms.



We can see that the equation {{{y=-2*x+4}}} has a slope {{{m=-2}}} and a y-intercept {{{b=4}}}.



Now to find the slope of the perpendicular line, simply flip the slope {{{m=-2}}} to get {{{m=-1/2}}}. Now change the sign to get {{{m=1/2}}}. So the perpendicular slope is {{{m=1/2}}}.



Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=-2}}} and the coordinates of the given point *[Tex \LARGE \left\(6,7\right\)].



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



{{{y-7=(1/2)(x-6)}}} Plug in {{{m=1/2}}}, {{{x[1]=6}}}, and {{{y[1]=7}}}



{{{y-7=(1/2)x+(1/2)(-6)}}} Distribute



{{{y-7=(1/2)x-3}}} Multiply



{{{y=(1/2)x-3+7}}} Add 7 to both sides. 



{{{y=(1/2)x+4}}} Combine like terms. 



So the equation of the line perpendicular to {{{2x+y=4}}} that goes through the point *[Tex \LARGE \left\(6,7\right\)] is {{{y=(1/2)x+4}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-2*x+4,(1/2)x+4)
circle(6,7,0.08),
circle(6,7,0.10),
circle(6,7,0.12))}}}Graph of the original equation {{{y=-2*x+4}}} (red) and the perpendicular line {{{y=(1/2)x+4}}} (green) through the point *[Tex \LARGE \left\(6,7\right\)]. 




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# 3



{{{8x+7y=3}}} Start with the given equation.



{{{7y=3-8x}}} Subtract {{{8x}}} from both sides.



{{{7y=-8x+3}}} Rearrange the terms.



{{{y=(-8x+3)/(7)}}} Divide both sides by {{{7}}} to isolate y.



{{{y=((-8)/(7))x+(3)/(7)}}} Break up the fraction.



{{{y=-(8/7)x+3/7}}} Reduce.



We can see that the equation {{{y=-(8/7)x+3/7}}} has a slope {{{m=-8/7}}} and a y-intercept {{{b=3/7}}}.



Now to find the slope of the perpendicular line, simply flip the slope {{{m=-8/7}}} to get {{{m=-7/8}}}. Now change the sign to get {{{m=7/8}}}. So the perpendicular slope is {{{m=7/8}}}.



Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=-8/7}}} and the coordinates of the given point *[Tex \LARGE \left\(7,-8\right\)].



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



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



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



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



{{{y+8=(7/8)x-49/8}}} Multiply



{{{y=(7/8)x-49/8-8}}} Subtract 8 from both sides. 



{{{y=(7/8)x-113/8}}} 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>.



So the equation of the line perpendicular to {{{8x+7y=3}}} that goes through the point *[Tex \LARGE \left\(7,-8\right\)] is {{{y=(7/8)x-113/8}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-(8/7)x+3/7,(7/8)x-113/8)
circle(7,-8,0.08),
circle(7,-8,0.10),
circle(7,-8,0.12))}}}Graph of the original equation {{{y=-(8/7)x+3/7}}} (red) and the perpendicular line {{{y=(7/8)x-113/8}}} (green) through the point *[Tex \LARGE \left\(7,-8\right\)]. 


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# 4


This problem is identical to the first question. Make sure that you've entered everything correctly



<hr>


# 5



{{{5x=7y+9}}} Start with the given equation.



{{{5x-9=7y}}} Subract 9 from both sides.



{{{(5x-9)/7=y}}} Divide both sides by 7 to isolate y.



{{{y=(5x-9)/7}}} Rearrange the equation.



{{{y=(5/7)x-9/7}}} Break up the fraction.




We can see that the equation {{{y=(5/7)x-9/7}}} has a slope {{{m=5/7}}} and a y-intercept {{{b=-9/7}}}.



Since parallel lines have equal slopes, this means that we know that the slope of the unknown parallel line is {{{m=5/7}}}.



Now let's use the point slope formula to find the equation of the parallel line by plugging in the slope {{{m=5/7}}}  and the coordinates of the given point *[Tex \LARGE \left\(-9,7\right\)].



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



{{{y-7=(5/7)(x--9)}}} Plug in {{{m=5/7}}}, {{{x[1]=-9}}}, and {{{y[1]=7}}}



{{{y-7=(5/7)(x+9)}}} Rewrite {{{x--9}}} as {{{x+9}}}



{{{y-7=(5/7)x+(5/7)(9)}}} Distribute



{{{y-7=(5/7)x+45/7}}} Multiply



{{{y=(5/7)x+45/7+7}}} Add 7 to both sides. 



{{{y=(5/7)x+94/7}}} 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>.



So the equation of the line parallel to {{{5x=7y+9}}} that goes through the point *[Tex \LARGE \left\(-9,7\right\)] is {{{y=(5/7)x+94/7}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,(5/7)x-9/7,(5/7)x+94/7),
circle(-9,7,0.08),
circle(-9,7,0.10),
circle(-9,7,0.12))}}}Graph of the original equation {{{5x=7y+9}}} (red) and the parallel line {{{y=(5/7)x+94/7}}} (green) through the point *[Tex \LARGE \left\(-9,7\right\)].