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# help

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The graph of the line $x+y=b$ is a perpendicular bisector of the line segment from $(1,3)$ to $(5,7)$. What is the value of b?

gueesstt  May 4, 2018
#1
+20153
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The graph of the line $x+y=b$ is a perpendicular bisector of the line segment from $(1,3)$ to $(5,7)$.

What is the value of b?

$$\begin{array}{|lrcll|} \hline & \vec{x} &=& \dbinom{ \frac{1+5}{2} } { \frac{3+7}{2} } + \lambda \binom{ -(7-3) } {5-1 }\\ & &=& \dbinom{3} { 5 } + \lambda \binom{ -4 } { 4 }\\ & \dbinom{x}{y} &=& \dbinom{3-4\lambda}{5+4\lambda} \\\\ (1) & x &=& 3-4\lambda \\ (2) & y &=& 5+4\lambda \\\\ (1)+(2): & x+y &=& 3-4\lambda +5+4\lambda \\ & x+y &=& 3 +5 \\ & x+y &=& 8 \quad & | \quad x+y=b\\ & &&& | \quad \mathbf{b = 8} \\ \hline \end{array}$$

heureka  May 4, 2018
#2
+91148
+3

The midpoint  of this segment is  ( 3, 5)

And the slope between the two segment endpoints  is  [7 - 3] / [ 5 - 1]   = 4/4  = 1

So.......the perpedicular bisector  will have a negative reciprocal slope  = -1

And the equation of this bisector is given  by :

y = -1(x - 3) + 5

y = -x + 3 + 5

y = -x + 8

So

x + y  =  8

And  "b"  =  8

CPhill  May 4, 2018