Question:

A light ray emerging from the point source placed at $P( 1,3)$ is reflected at a point $Q$ in the axis of $x$. If the reflected ray passes through the point $R (6, 7)$, then the abscissa of $Q$ is:

Updated On: Oct 14, 2024
  • $1$
  • $3$
  • $\frac{7}{2}$
  • $\frac{5}{2}$
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The Correct Option is D

Solution and Explanation

Let abcissa of $Q = x$
$\therefore Q = \left(x, 0\right)$
$tan\,\theta = \frac{0-7}{x-6}, tan \left(180^{\circ}-\theta\right) = \frac{0-3}{x-1}$
Now, tan $\left(180^{\circ}- \theta\right)=- tan \,\theta$
$\therefore\quad \frac{-3}{x-1} = \frac{-7}{x-6} \Rightarrow x = \frac{5}{2}$
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Questions Asked in JEE Main exam

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Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points. 

A straight line can be represented as an equation in various forms,  as show in the image below:

 

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y0 = m (x – x0)

2. Two – Point Form

Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2)  are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes

The slope of P2P = The slope of P1P2 , i.e.

\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)

Hence, the equation becomes:

y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c