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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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What are the points on the $y$ - axis whose distance from the line $ \large\frac{x}{3}$$+ \large\frac{y}{4}$$=1$ is 4 units.

$\begin {array} {1 1} (A)\;\bigg(0, \large\frac{32}{3} \bigg)\: and \: \bigg( 0, -\large\frac{8}{3} \bigg) & \quad (B)\;\bigg( 0, -\large\frac{32}{3} \bigg) \; and \: \bigg( 0, \large\frac{8}{3} \bigg) \\ (C)\;\bigg( 0, -\large\frac{32}{3} \bigg) \; and \: \bigg( 0,- \large\frac{8}{3} \bigg) & \quad (D)\;\bigg( 0, \large\frac{32}{3} \bigg) \; and \: \bigg( 0, \large\frac{8}{3} \bigg) \end {array}$

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  • Perpendicular distance of a line $Ax+By+C=0$ from a point $(x_1, y_1)$ is given by $ d = \large\frac{|Ax_1+By_1+C|}{\sqrt {A^2+B^2}}$
Let (o, b) be the point on the $y$ - axis given that the distance from the line
$ \large\frac{x}{3}$$+\large\frac{y}{4}$$=1$ is 4 units
This can be written as $ 4x+3y-12=0$--------(1)
Comparing this equation with the general equation
$Ax+By+C=0$ we get,
$A=4, B=3$ and $C = -12$
If (o, b) is the point on the $y$ - axis, then $d = \large\frac{|Ax_1+By_1+C|}{\sqrt {A^2+B^2}}$
$ 4 = \large\frac{|4(0)+3(b)-12|}{\sqrt{4^2+3^2}}$
$ \Rightarrow 4 = \large\frac{|3b-12|}{\sqrt{25}}$$ = \large\frac{|3b-12|}{5}$
$ \therefore 20 = \pm (3b-12)$
Case (i)
when 20 = + (3b-12)$
$ \Rightarrow 3b = 20+12$
$ \Rightarrow b = \large\frac{32}{3}$
Case (ii)
when 20 = -(3b-12)$
$ \Rightarrow 3b=-8$
$ \Rightarrow b = -\large\frac{8}{3}$
Hence the required points are
$\bigg(0, \large\frac{32}{3} \bigg)$ and $\bigg(0, -\large\frac{8}{3} \bigg)$
answered May 18, 2014 by thanvigandhi_1
 

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