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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, – 1)

$\begin {array} {1 1} (A)\;x−4y+3=0 & \quad (B)\;x+4y+3=0 \\ (C)\;x−4y−3=0 & \quad (D)\; x+4y−3=0 \end {array}$

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  • Equation of a line pasing through two points $(x_1,y_1)$ and $(x_2, y_2)$ is $ \large\frac{y-y_1}{y_2-y_1}$$ = \large\frac{x-x_1}{x_2-x_1}$
  • If two lines are perpendicular, then the product of their slopes is -1.
Step 1
Equation of the line passing through the points (2, 3) and (3, -1 ) is
$ \large\frac{y-3}{-1-3}$$= \large\frac{x-2}{3-2}$
$ \Rightarrow \large\frac{y-3}{-4}$$=\large\frac{x-2}{1}$
or $y-3=-4(x-2)$
$ \Rightarrow y-3 = -4x+8$
(i.e.,) $4x+y=11$
This is the equation of the line.
Step 2
The line which is perpendicular to this line is $ x-4y+k=0$
Since it passes through the point (5,2)
$ 5-4(2)+k=0$
$ \Rightarrow k-3=0$
or = k = 3$
Hence the equation of the required line
$ x -4y+3=0$
answered Jun 30, 2014 by thanvigandhi_1

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