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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Choose the correct answer from the given four options. A point equidistant from the lines $4x + 3y + 10 = 0, 5x – 12y + 26 = 0$ and $7x+24y-50=0$ is

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

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1 Answer

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  • Distance of a line $ax+by+c=0$ from a point $(x_1,y_1)$ is $ d = \bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$
Step 1 :
Consider the point (1,-1).
Substituting for x and y in the equations we get,
$d_1 = \bigg| \large\frac{4(1)+3(-1)+10}{\sqrt{4^2+3^2}} \bigg|$
$ = \large\frac{11}{5}$
$d_2 = \bigg| \large\frac{5(1)+(-12)(-1)+26}{\sqrt{25+144}} \bigg|$
$ = \large\frac{43}{13}$
$d_3 = \bigg| \large\frac{7(1)+24(-1)-50}{\sqrt{49+576}} \bigg|$
$ = \large\frac{64}{25}$
Hence $ d_1 \neq d_2 \neq d_3$
Consider the point (1,1)
$ \therefore d_1 = \bigg| \large\frac{4(1)+3(1)+10}{\sqrt{4^2+3^2}} \bigg|$
$ = \large\frac{17}{25}$
$d_2 = \bigg| \large\frac{5(1)-12(1)+26}{\sqrt{25+144}} \bigg|$
$ = \large\frac{19}{13}$
$d_3 = \bigg| \large\frac{7(1)+24(1)-50}{\sqrt{7^2+24^2}} \bigg|$
$ = \large\frac{19}{25}$
$ \therefore d_1 \neq d_2 \neq d_3$
Consider the point (0,0)
$ d_1 = \bigg| \large\frac{4(0)+3(0)+10}{\sqrt{4^2+3^2}} \bigg|$
$ = \large\frac{10}{5}$ = 2 units.
$d_2 = \bigg| \large\frac{5(0)-12(0)+26}{\sqrt{25+144}} \bigg|$
$ = \large\frac{26}{13}$ = 2 units
$d_3 = \bigg| \large\frac{7(0)+24(0)-50}{\sqrt{7^2+24^2}} \bigg|$
$ = \large\frac{50}{25}$ = 2 units
It is clear that $d_1=d_2=d_3$
Hence 'C' is the correct options.
answered Jul 6, 2014 by thanvigandhi_1
 

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