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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Fill in the blank for the following : A point moves so that square of its distance from the point (3, –2) is numerically equal to its distance from the line $ 5x – 12y = 3$ . The equation of its locus is ____.

$\begin {array} {1 1} (A)\;13(x^2+y^2)+83x-64y-172=0 & \quad (B)\;13(x^2+y^2)-83x+64y-172=0 \\ (C)\;13(x^2+y^2)-83x+64y+172=0 & \quad (D)\;13(x^2+y^2)-83x-64y+172=0 \end {array}$

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1 Answer

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Toolbox:
  • Distance between a line $ax+by+c$ from the point $(x_1, y_1)$ is $ d = \bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$
  • Distance between two points $(x_1, y_1) $ and $(x_2, y_2)$ is $d = \sqrt {(x_2-x_1)^2+(y_2-y_1)^2}$
Step 1 :
Equation of the given line is $5x-12y-3$.
Let the point be $(x_1, y_1)$
Hence the distance of this line from the point $(x_1, y_1)$ is
$ d = \bigg| \large\frac{5x_1-12y_1-3}{\sqrt{5^2+12^2}} \bigg|$$ = \large\frac{5x_1-12y_1-3}{13}$
Also the distance between the points $(x_1, y_1)$ and (3, -2) is
$ d = \sqrt{(x_1-3)^2+(y_1+2)^2}$
According to the given conditions
$( \sqrt{(x_1-3)^2+(y_1+2)^2} )^2$ = $ \large\frac{(5x_1-12y_1-3)}{13} $
(i.e) $x_1^2-6x_1+9+y_1^2+4y_1+4 = \large\frac{(5x-12y-3)}{13}$
$ \Rightarrow 13(x_1^2-6x_1+9 + y_1^2+4y_1+4)= 5x-12y-3$
$ \Rightarrow 13(x_1^2+y_1^2)-83x_1+64y_1+172=0$
Hence the equation of the locus of the point is $13(x^2+y^2)-83x+64y+172=0$
answered Jul 8, 2014 by thanvigandhi_1
 

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