Question

# The slope of a line joining $$P(6, k)$$ and $$Q(1 - 3k, 3)$$ is $$\dfrac {1}{2}$$. Find midpoint of $$PQ$$

Solution

## Let $$P(6, k) = (x_{1}, y_{1})$$ and $$Q(1 - 3k, 3) = (x_{2}, y_{2})$$$$\therefore$$ Slope of $$PQ = \dfrac {y_{2} - y_{1}}{x_{2} - x_{1}}$$$$\Rightarrow \dfrac {3 - k}{1 - 3k - 6} = \dfrac {1}{2}$$$$6 - 2k = -5 - 3k$$$$11 = -k$$$$k = -11$$Hence, $$P \equiv \left(6, -11 \right)$$ and $$Q \equiv \left(1-(3\times -11), 3\right) \equiv (34,3)$$Midpoint of $$PQ \equiv \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}2 \right) \equiv \left(\dfrac{6+34}2, \dfrac{-11+3}2\right)\equiv (20,-4)$$Mathematics

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