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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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$A(1,2)$ & $B(7,6)$ are the two opposite points of straight line (i.e) these two points are on opposite side of a straight line then find a equation of straight line which is equidistant from these two points and also AB is perpendicular to the straight line.

$\begin{array}{1 1}(a)\;2y+3x=20\\(b)\;3y+2x=20\\(c)\;y+3x=10\\(d)\;2y+x=20\end{array}$

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1 Answer

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Mid-point of the points $A(1,2)$ & $B(7,6)$ is $C(x,y)$
$x=\large\frac{1+7}{2}$
$x=4$
$y=\large\frac{2+6}{2}$
$y=4$
Hence straight line passes through (4,4)
BC is perpendicular to straight line.Hence slope of $BC=\large\frac{6-4}{7-4}=\frac{2}{3}$
So the slope of straight line will be $-\large\frac{3}{2}$
Hence equation of straight line is $(y-4)=-\large\frac{3}{2}$$(x-4)$
$2y-8=-3x+12$
$2y+3x=20$
Hence (a) is the correct answer.
answered Feb 4, 2014 by sreemathi.v
 

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