logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Straight Lines
0 votes

A ray of light passing through the point (1, 2) reflects on the $x$ - axis at point $A$ and the reflected ray passes through the point (5, 3). Find the coordinates of $A$

$\begin {array} {1 1} (A)\;\bigg( \large\frac{5}{13}, 0 \bigg) & \quad (B)\;\bigg( -\large\frac{5}{13}, 0 \bigg) \\ (C)\;\bigg( \large\frac{13}{5}, 0 \bigg) & \quad (D)\;\bigg( -\large\frac{13}{5}, 0 \bigg) \end {array}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Slope of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $ m=\large\frac{y_2-y_1}{x_2-x_1}$
  • Also slope $m=\tan \theta$ where $\theta$ is the angle of inclination.
Let the coordinates of the point A be ( a,0) .
Let AL be the perpendicular to $x$ - axis.
By Law of reflection we know,
angle of incidence = angle of reflection.
Hence $ < BAL = < CAL = \phi$
Let $ < CAX = \theta$
$ < OAB = 180^{\circ} - ( \theta + 2\phi)$
$ \qquad \qquad = 180^{\circ} - [ \theta + 2( 90^{\circ}- \theta )$
$( \because \phi = 90 - \theta)$
$ \therefore < OAB = 180^{\circ} - \theta - 180^{\circ} + 2\theta$
$ = \theta \Rightarrow < BAX = 180^{\circ} - \theta$
Slope of the line $AC, m = \large\frac{3-0}{5-a}$
also $m = \tan \theta$
$ \Rightarrow \tan \theta = \large\frac{3}{5-a}$-------(1)
Slope of the line $AB = \large\frac{2-0}{1-a}$
$ \therefore \tan ( 180^{\circ} - \theta ) = \large\frac{2}{1-a}$
But $ \tan ( 180^{\circ}-\theta)=-\tan \theta$
$ \therefore -\tan \theta = \large\frac{2}{1-a}$
$ \therefore \tan \theta = \large\frac{2}{a-1}$------(2)
equally (1) and (2) we get,
$ \large\frac{3}{5-a}$$ = \large\frac{2}{a-1}$
$ \Rightarrow 3(a-1)=2(5-a)$
$ 3a-3=10-2a$
$ \Rightarrow 5a=13$
$ \therefore a = \large\frac{13}{5}$
Hence the coordinate of A are $ \bigg( \large\frac{13}{5}$$, 0 \bigg)$
answered May 23, 2014 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...