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.

xplosions Jul 3, 2020

#1**-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.

Guest Jul 3, 2020

#3**+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?

!

asinus Jul 3, 2020