What is the equation of the right bisector of the line segment joining (1, 1) and (2, 3)?

2 x + 4 y - 11 = 0

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2 x - 4 y - 5 = 0

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2 x - 4 y - 11 = 0

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x - y + 1 = 0

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Solution

The correct option is C $2 x + 4 y - 11 = 0$
Let the points be $A (1, 1)$ and $B (2, 3)$
Slope of $A B = \frac{3 - 1}{2 - 1} = 2$
Product of slope of a line and its perpendicular is $- 1$
$\Rightarrow$ Slope of perpendicular $= - \frac{1}{2}$
Mid point of $A B$ is
$(\frac{1 + 2}{2}, \frac{1 + 3}{2}) (\frac{3}{2}, 2)$
So the equation of perpendicular is
$y - 2 = - \frac{1}{2} (x - \frac{3}{2}) 4 y - 8 = - (2 x - 3) 2 x + 4 y = 11 2 x + 4 y - 11 = 0$

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