Question 332266

Triangle ABC has vertices A (-5, 4), B (1, -2) and C (3, 6),
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That triangle is:

{{{drawing(400,400,-7,7,-7,7, triangle(-5, 4,1, -2,3, 6), graph(400,400,-7,7,-7,7), red(locate(-7,4,"A(-5,4)" ), locate(1.3,-2,"B(1,-2)"), locate(3,6,"C(3,6)"))

 )}}}
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(a)Write the equation of the line of which AB is a segment.
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That's the green line below:

{{{drawing(400,400,-7,7,-7,7, triangle(-5, 4,1, -2,3, 6), graph(400,400,-7,7,-7,7), red(locate(-7,4,"A(-5,4)" ), locate(1.3,-2,"B(1,-2)"), locate(3,6,"C(3,6)")),green(line(-9,8,8,-9))

 )}}}

{{{m=(y[2]-y[1])/(x[2]-x[1])=((-2)-(4))/((1)-(-5))=(-2-4)/(1+5)=-6/6=-1}}}

{{{y-y[1]=m(x-x[1])}}}

{{{y-(4)=(-1)(x-(-5))}}}

{{{y-4=-(x+5)}}}

{{{y-4=-x-5}}}

{{{y=-x-1}}}

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(b)Write the equation of the line of which the altitude to line AC is a segment.
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That's the blue line below:

{{{drawing(400,400,-7,7,-7,7, triangle(-5, 4,1, -2,3, 6), graph(400,400,-7,7,-7,7), red(locate(-7,4,"A(-5,4)" ), locate(1.3,-2,"B(1,-2)"), locate(3,6,"C(3,6)")), blue(line(8,-30, -8,34))

 )}}}

Since it is an altitude, it is perpendicular to AC.

First we find the slope of AC:

{{{m=(y[2]-y[1])/(x[2]-x[1])=((6)-(4))/((3)-(-5))=(6-4)/(3+5)=2/8=1/4}}}

The blue line is perpendicular to AC so its slope is
the reciprocal of {{{1/4}}} with the opposite sign, so the
blue line's slope is {{{-4}}} and it passes through B(1,-2)

{{{y-y[1]=m(x-x[1])}}}

{{{y-(-2)=(-4)(x-(1))}}}

{{{y+2=-4(x-1)}}}

{{{y+2=-4x+4}}}

{{{y=-4x+2}}}
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(c) Write the equation of the perpendicular bisector of line AC.
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We find the midpoint of AC, using the midpoint formula:

Midpoint = ({{{(x[1]+x[2])/2}}},{{{(y[1]+y[2])/2 )}}}) = ({{{((-5)+(3))/2}}},{{{(4+6)/2 )}}}) = ({{{(-2)/2}}},{{{(10)/2 )}}}) = M(-1,5)

Now we find the equation of the red line through M(-1,5), it has the same 
slope as the altitude since both are perpendicular to the line AC.

{{{drawing(400,400,-7,7,-7,7, triangle(-5, 4,1, -2,3, 6), graph(400,400,-7,7,-7,7), red(locate(-7,4,"A(-5,4)" ), locate(1.3,-2,"B(1,-2)"), locate(3,6,"C(3,6)")), blue(line(8,-30, -8,34)), red(line(-9,37,9,-35)),
locate(-3,5,"M(-1,5)")

 )}}}

The red line's slope is also {{{-4}}} but it passes through M(-1,5)

{{{y-y[1]=m(x-x[1])}}}

{{{y-(5)=(-4)(x-(-1))}}}

{{{y-5=-4(x+1)}}}

{{{y-5=-4x-4}}}

{{{y=-4x+1}}}

{{{drawing(400,400,-7,7,-7,7, triangle(-5, 4,1, -2,3, 6), graph(400,400,-7,7,-7,7), red(locate(-7,4,"A(-5,4)" ), locate(1.3,-2,"B(1,-2)"), locate(3,6,"C(3,6)")), blue(line(8,-30, -8,34)), red(line(-9,37,9,-35)),
locate(-3,5,"M(-1,5)")

 )}}}
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(d)Find the perimeter of triangle ABC.
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We use the distance formula to find the lengths of all three sides.

{{{d=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

{{{AB=sqrt( ((1)-(-5))^2+((-2-(4))^2 ))=sqrt((1+5)^2+(-2-4)^2)=sqrt(6^2+(-6)^2)=sqrt(36+36)= sqrt(72)=sqtr(36*2)=6sqrt(2)}}}

{{{BC=sqrt( ((3)-(1))^2+((6-(-2))^2 ))=sqrt((3-1)^2+(6+2)^2)=sqrt(2^2+8^2)=sqrt(4+64)= sqrt(68)=sqrt(4*17)=2sqrt(17)}}}

{{{AC=sqrt( ((3)-(-5))^2+(((6)-(4))^2 ))=sqrt((3+5)^2+(6-4)^2)=sqrt(8^2+2^2)=sqrt(64+4)= sqrt(68)=sqrt(4*17)=2sqrt(17)}}}

So the perimeter = AB + BC + AC = {{{6sqrt(2)+2sqrt(17)+2sqrt(17)=6sqrt(2)+ 4sqrt(17)}}}

Edwin</pre>