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The perpendicular bisector of the line segment connecting the points (-3,8) and (-5,4) has an equation of the form y=mx+b. Find m+b.

 Jul 3, 2020
 #1
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-2

The slope of the line segment is (8 - 4)/(-5 - (-3)) = -2, so the slope of the perpendicular bisector is 1/2.  The midpoint is (-4,6), so by point-slope form, the equation fo the line is y - 6 = 1/2(x + 4).  Then y = 1/2*x + 8, so m + b = 1/2 + 8 = 17/2.

 Jul 3, 2020
 #2
avatar+107 
-2

guest your answer is wrong

 Jul 3, 2020
 #3
avatar+10390 
+1

huh im confused and its due today

The perpendicular bisector of the line segment connecting the points (-3,8) and (-5,4) has an equation of the form y=mx+b. Find m+b.

 

Hello xplosions!

 

Die senkrechte Halbierende des Liniensegments, das die Punkte (-3,8) und (-5,4) verbindet, hat eine Gleichung der Form y = mx + b. Finde m + b.

 

Two point equation

\(f_1(x)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1\)

\(f_1(x)=\frac{4-8}{(-5)-(-3)}(x-(-3))+8\\ f_1(x)=\frac{-4}{-2}(x+3)+8\)

\(f_1(x)=2x+14\)

\(m_1=2\\ b_1=14\)

 

\(P_2(\frac{-3+(-5)}{2},\frac{8+4}{2})\)

\(P_2(-4,6)\)

\(m_2=-\frac{1}{m_1}\)

\( m_2=-\frac{1}{2}\)

Point direction equation

\(f_2(x)=m_2(x-x_2)+y_2\)

\(f_2(x)=-\frac{1}{2}(x-(-4))+6\)

\(f_2(x)=-\frac{1}{2}x+4 \)

\(b_2=4\)

 

\(m_2+b_2=\frac{7}{2}\)            Wozu braucht man diese Addition?

laugh  !

 Jul 3, 2020
 #4
avatar+107 
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Thank you!!! I don't actually know what language you speak though...

 Jul 3, 2020
 #5
avatar+10390 
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German

Wozu braucht man diese Addition?   >>>  Why do you need this addition?

Danke für dein Danke. It doesn't happen that often here.

laugh  !

asinus  Jul 3, 2020
edited by asinus  Jul 3, 2020
 #6
avatar+1129 
0

So for perpendicular bisectors has slope 1/6 since the first one has slope -6 and we then get thta the y-intercept is 3 1/3 so we het that the answer is 7/2.

 Jul 3, 2020
edited by jimkey17  Jul 3, 2020

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