Question

A line forms a triangle of area $54\sqrt 3$ sq. with coordinate axes. Find the equation of the line if $\bot$ from origin to the line makes an angle of $60^\circ$ with x-axis.

Answer 1

As mentioned in the question the line forms a triangle with coordinate axes which clearly means that the triangle will be a right-angled triangle. The Area of the triangle is also given so we need to calculate the height and base to form the required equation.

Complete step-by-step answer:

In the question, it is given that a line forms a triangle of area $54\sqrt 3$ sq. with coordinate axes,
So, let us take a line $AB$ as shown in figure which forms a triangle of area $54\sqrt 3$ sq. with coordinate axes.
Since, we can see that point $A$ lies on x-axis, therefore the y-coordinate of $A$ will be $0$
So, $A = \left( {a,0} \right)$
Also, we can see that point $B$ lies on y-axis, therefore the x-coordinate of $B$ will be $0$
So, $B = \left( {0,b} \right)$
As we know, area of a triangle $= \dfrac{1}{2} \times height \times base$
Now, in $\vartriangle AOB$ , height $= b$ and base $= a$
Area of $\vartriangle AOB = \dfrac{1}{2} \times b \times a$
And Area of $\vartriangle AOB$ = $54\sqrt 3$
Therefore, $54\sqrt 3 = \dfrac{1}{2}ab$
Or, $ab = 108\sqrt 3$ --- $\left( 1 \right)$
Now, the perpendicular drawn from the origin to the line makes an angle of $60^\circ$ with x-axis as shown in figure.
Let the length of the perpendicular drawn from origin to line be $P$ .
Also, we know that in a right-angled triangle, $\cos \theta = \dfrac{{Base}}{{Hypotenuse}}$
$\cos 60^\circ = \dfrac{P}{a}$ and $\cos 60^\circ = \dfrac{1}{2}$
Or, $a = \dfrac{P}{{\cos 60^\circ }} = 2P$
And, $\cos 30^\circ = \dfrac{P}{b}$ and $\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}$
Or, $b = \dfrac{P}{{\cos 30^\circ }} = \dfrac{{2P}}{{\sqrt 3 }}$
Now, put values of $a$ and $b$ in equation $\left( 1 \right)$ ,
$2P \times \dfrac{{2P}}{{\sqrt 3 }} = 108\sqrt 3$
Or, $\dfrac{{4{P^2}}}{{\sqrt 3 }} = 108\sqrt 3$
Or, ${P^2} = 108 \times 3$
$P = \pm 18$
But we can only take $P = 18$ because the triangle is in the${1^{st}}$ quadrant.
So, $P = 18$
And, $a = 2P = 36$ , $b = \dfrac{{2P}}{{\sqrt 3 }} = \dfrac{{36}}{{\sqrt 3 }} = 12\sqrt 3$
The intercept form of the equation of the straight line is $\dfrac{x}{a} + \dfrac{y}{b} = 1$
So, the equation of the line becomes $\dfrac{x}{{36}} + \dfrac{y}{{12\sqrt 3 }} = 1$
Hence, the equation of line comes out to be $x + \sqrt 3 y - 36 = 0$

Note: Few key points used in this question which needs to be remembered are- the points where
line cuts the coordinate axes are called intercepts, area of right-angled triangle is $\dfrac{1}{2} \times height \times base$
, and the intercept form of the equation of the straight line is $\dfrac{x}{a} + \dfrac{y}{b} = 1$