The coordinates of A and B are -3- and 1-4 and the mid point of AB is P-1-1- Find the value of -

A=( 3,α) #160; #160; #160; B=(1,α +4) Mid point of AB is P( 1,1) 3+1 2 = 2 2 = 1 ; α +α +4 2 =1 2α +4=2 ...

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Question

The coordinates of $A$ and $B$ are $(- 3, α)$ and $(1, α + 4)$ and the mid point of $¯ ¯¯¯¯¯¯ ¯ A B$ is $P (- 1, 1)$ . Find the value of $α$

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Solution

$A = (- 3, α)$ $B = (1, α + 4)$
Mid point of $A B$ is $P (- 1, 1)$
$\frac{- 3 + 1}{2} = \frac{- 2}{2} = - 1; \frac{α + α + 4}{2} = 1 2 α + 4 = 2 \Rightarrow 2 α = - 2 \Rightarrow α = - 1$

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The coordinates of A and B are (−3,α) and (1,α+4) and the mid point of ¯¯¯¯¯¯¯¯AB is P(−1,1). Find the value of α

A=(−3,α) B=(1,α+4)Mid point of AB is P(−1,1)−3+12=−22=−1;α+α+42=12α+4=2⇒2α=−2⇒α=−1

The coordinates of $A$ and $B$ are $(- 3, α)$ and $(1, α + 4)$ and the mid point of $¯ ¯¯¯¯¯¯ ¯ A B$ is $P (- 1, 1)$ . Find the value of $α$

$A = (- 3, α)$ $B = (1, α + 4)$
Mid point of $A B$ is $P (- 1, 1)$
$\frac{- 3 + 1}{2} = \frac{- 2}{2} = - 1; \frac{α + α + 4}{2} = 1 2 α + 4 = 2 \Rightarrow 2 α = - 2 \Rightarrow α = - 1$