Question:

The minimum area of the triangle formed by the variable line $3 \, \cos\,\theta \cdot x + 4 \,\sin \,\theta \cdot y = 12$ and the co-ordinate axes is

Updated On: Feb 21, 2024
  • $144$
  • $\frac {25}{2}$
  • $\frac {49}{4}$
  • $12$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

Given equation of line is
$x \cdot 3 \cos \theta+4 \,\sin \theta\, y=12$
$\Rightarrow \frac{x}{(4 / \cos \theta)}+\frac{y}{(3 / \sin \theta)}=1\,\,\,\,\,\,..(i)$
It intereset the coordinate axes at $A\left(\frac{4}{\cos \theta}, 0\right)$ and
$B\left(0, \frac{3}{\sin \theta}\right)$
$\therefore$ Area of $\Delta O A B$
$\Delta =\frac{1}{2} \times \frac{4}{\cos \theta} \times \frac{3}{\sin \theta} $
$=\frac{12}{\sin 2 \,\theta}\,\,\,\,\,\,\,\,..(ii)$
Now, for area to be minimum,
$\sin 2\, \theta$ should be maximum i.e.,
$\sin 2 \,\theta=1$
$\sin 2 \,\theta \mid \leq 1)\,\,\,\,\,\,\,\,\,(\because|\sin 2 \theta| \leq 1)$
$\therefore$ Minimum area
$\Delta_{\min }=\frac{12}{1}=12$
Was this answer helpful?
1
0

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