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Find the points on the $x$ axis, whose distance from the line $\large\frac{x}{3}$$+\large\frac{y}{4}$$=1$ are 4 units.

$\begin{array}{1 1}(A)\;(-2,8)(0,0) \\(B)\; (2,0)(8,0) \\(C)\; (-2,0)(8,0) \\(D)\;(2,8)(0,0) \end{array} $

1 Answer

  • Distance of a point $(x_1,y_1)$ from the given line is $ d=\large\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
The given equation of the line is
$ \large\frac{x}{3}$$+\large\frac{y}{4}$$=1$
This can be written as $4x+3y-12=0$
The general form of an equation of a straight line is $Ax+By+C=0$
The perpendicular distance $(d)$ of the line from a point $(x_1, y_1)$ is
$ d = \large\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
Let (a, o)be the point on the $x$ - axis.
Comparing both the equations we get,
$A=4, B=3\: and \: C=-12$ and given $d = 4$ units.
Substituting the values we get,
$ 4 = \large\frac{|4a+3x0-12|}{\sqrt{4^2+3^2}}$
$ \Rightarrow 4 = \large\frac{|4a-12|}{\sqrt{16+9}}$$ = \large\frac{|4a-12|}{5}$
$ \Rightarrow 20=|4a-12|$
Case (i) $\quad 20=4a-12$
$ \qquad \quad \Rightarrow 4a=32$
$\qquad \quad \therefore a=8$
Case (ii) $\quad 20=-(4a-12)$
$ \qquad \quad \Rightarrow 4a=-8$
$\qquad \quad \therefore a=-2$
Hence the required points are (-2,0) and (8,0)
answered May 9, 2014 by thanvigandhi_1

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