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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the distance of the line $4x + 7y + 5 = 0$ from the point (1, 2) along the line $2x-y=0$

$\begin {array} {1 1} (A)\;\large\frac{5\sqrt{23}}{18} units & \quad (B)\;\large\frac{18\sqrt{23}}{5} units \\ (C)\;\large\frac{23\sqrt{5}}{18} units & \quad (D)\;\large\frac{5}{18} units \end {array}$

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  • Distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $ \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
The given lines are
$ \qquad 2x-y=0$ ------(1)
$ \qquad 4x+7y+5=0$-----(2)
Given point on line (1) is $A(1,2)$
Let the point of intersection of the two lines be B.
Let us now find the point of intersection of the two lines.
$ \qquad ( \times 2) 2x-y=0$
$\qquad \quad \: \: 4x+7y=-5$
$ \qquad 4x-2y=0$
$ \qquad 4x+7y=-5$
$ \qquad \qquad -9y=5 \Rightarrow y = -\large\frac{5}{9}$
Substituting the value of $y$ in (1) we get,
$2x- \bigg( -\large\frac{5}{9} \bigg) = 0$
$ \Rightarrow 2x=-\large\frac{5}{9}$ or $ x= \large\frac{-5}{18}$
The coordinate of point B is $ \bigg( \large\frac{-5}{18}$$, \large\frac{-5}{9} \bigg)$
Hence the distance between the two points $A(1,2)$ and $B \bigg( \large\frac{-5}{18}$$, \large\frac{-5}{9} \bigg)$
$ \overline {AB} = \sqrt{\bigg( 1+ \large\frac{5}{18} \bigg)^2+ \bigg( 2 + \large\frac{5}{9} \bigg)^2}$ units.
$ = \sqrt{ \bigg( \large\frac{23}{18} \bigg)^2 + \bigg( \large\frac{23}{9} \bigg)^2}$ units
$ = \sqrt{ \bigg( \large\frac{23}{9} \bigg)^2 \bigg( \large\frac{1}{4}+1 \bigg)}$ units
$ = \large\frac{23}{9} \sqrt{ \large\frac{5}{4}}$ units
$ = \large\frac{23 \sqrt 5}{18}$ units
Hence the required distance is $ \large\frac{23 \sqrt 5}{18}$ units.
answered May 21, 2014 by thanvigandhi_1

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