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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Choose the correct answer from the given four options. The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is

$\begin {array} {1 1} (A)\;\large\frac{130}{17\sqrt{29}} & \quad (B)\;\large\frac{13}{7\sqrt{29}} \\ (C)\;\large\frac{130}{7} & \quad (D)\;\text{None of these} \end {array}$

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  • Distance of a point $(x_1,y_1)$ from the line $ax+by+c=0$ is $d=\bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$
Step 1 :
Let us find the point of intersection of the lines
$(3) 2x-3y+5=0$
$(2) 3x + 4y \qquad=0$
___________________
$ \not6 \not x-9y+15=0$
$ \not 6 \not x +8y \qquad = 0$
$ (-) \quad (-)$
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$ \qquad -17y+15=0$
$ \therefore y=\large\frac{15}{17}$
and $x = \large\frac{20}{17}$
Hence the point of intersection is $ \bigg( \large\frac{20}{17}$$, \large\frac{15}{17} \bigg)$
Hence the distance of the line $5x-2y=0$ from the above point is
$ d = \bigg| \large\frac{5\bigg( \Large\frac{20}{17} \bigg)-2 \bigg( \Large\frac{15}{17} \bigg)}{\sqrt{5^2+2^2}} \bigg|$
$ = \bigg| \large\frac{\Large\frac{100}{17} - \Large\frac{30}{17}}{\sqrt{29}} \bigg|$
$ = \large\frac{130}{17\sqrt{29}}$
Hence option A is the correct answer.
answered Jul 3, 2014 by thanvigandhi_1
 

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