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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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The line through the points $(h,3)$ and $(4,1)$ intersects the line $7x-9y-19=0$ at right angle. Find the value of $h$.

$\begin{array}{1 1}(A)\;h=-\large\frac{22}{9} \\(B)\; h=\large\frac{22}{9} \\(C)\; h=\large\frac{9}{22} \\(D)\;h=-\large\frac{9}{22} \end{array} $

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  • Slope of a line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ is
  • $ m = \large\frac{y_2-y_1}{x_2-x_1}$
  • If two lines are perpendicular, then the product of their slopes is -1.
Slope of the line passing through $(h,3)$ and $(4,1)$ is
Equation of the given line is $ 7x-9y-19=0$
This can be written as
$ y = \large\frac{7}{9}$$x-\large\frac{19}{9}$
Hence the slope is $ \large\frac{7}{9}$$ \therefore m_2 = \large\frac{7}{9}$.
Since the lines are perpendicular, their product of the slopes is -1.
(i.e.,) $m_1m_2=-1$
$ \Rightarrow \large\frac{-2}{4-h}$$ \times \large\frac{7}{9}$$=-1$
$ \large\frac{-14}{36-9h}$$=-1$
$ \Rightarrow -14 = -36+9h$
$ \therefore 9h=22$
$ \therefore h = \large\frac{22}{9}$
answered May 10, 2014 by thanvigandhi_1

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