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Find the equation of the right bisector of the line segment joining the points (3,4) and (-1,2)

$\begin{array}{1 1}(A)\;2x+y=5 \\(B)\; 2x-y=5 \\(C)\; 2y-x=5 \\(D)\;2y+x=5 \end{array} $

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  • The coordinates of the midpoint of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $ \bigg( \large\frac{x_1+x_2}{2}$$, \large\frac{y_1+y_2}{2} \bigg)$
  • Slope of the line is $ \bigg( \large\frac{y_2-y_1}{x_2-x_1} \bigg)$
  • If two lines are perpendicular then the product of the slopes is -1.
The end points of the line segment are $A(3,4)$ and $B(-1,2)$
Hence the midpoint of the line AB is $ \bigg( \large\frac{3-1}{2}$$, \large\frac{4+2}{2} \bigg)$
$ \qquad = (1,3)$
Now the slope of the line is
$ m_1 = \large\frac{2-4}{-1-3}$$ = \large\frac{-2}{-4}$$ = \large\frac{1}{2}$
Hence the slope of the perpendicular will be
$m_2 = -\large\frac{1}{\bigg(\Large\frac{1}{2} \bigg)}$$=-2$
$\therefore $ Equation of the line passing through (1,3) and slope -2 is
(i.e., ) $(y-3)=-2x+2$
$ \Rightarrow 2x+y=5$
Hence the required equation of the line is $ 2x+y=5$
answered May 13, 2014 by thanvigandhi_1

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