Question

The equation of the lines on which the perpendiculars from the origin make \[{30^o}\] angle with x-axis and which form a triangle of area \[\dfrac{{50}}{{\sqrt 3 }}\] with axes, are
A. \[x + \sqrt 3 y \pm 10 = 0\]
B. \[\sqrt 3 x + y \pm 10 = 0\]
C. \[x \pm \sqrt 3 y - 10 = 0\]
D. None of these

Answer 1

Hint: As stated in the question, the line forms a triangle with coordinate axes, implying that the triangle will be right-angled. Because the area of the triangle is also given, we must compute the height and base to complete the necessary equation.

Formula Used: Triangle’s area or dimensions can be calculated using the formula
\[\dfrac{1}{2} \times {\rm{height}} \times {\rm{base}}\]

Complete step by step solution: We have been provided in the question that,
The perpendiculars from the origin make \[{30^o}\] angle with x-axis and which form a triangle of area \[\dfrac{{50}}{{\sqrt 3 }}\] with axes
And we can write the line’s equation \[AB\] as
\[x\cos \alpha + y\sin \alpha = p\]
We have been given that the perpendiculars from the origin make angle as \[{30^o}\]
So, we have
$x\cos30^\circ+y\sin30^\circ=p$
Now, we have to use trigonometry identity to write the values, we get
\[ \Rightarrow \dfrac{{\sqrt 3 x}}{2} + \dfrac{y}{2} = p\]
On simplification, we get
\[ \Rightarrow \sqrt 3 x + y = 2p \ldots (1)\]
Now we can see that in triangles OLA and OLB, we have
$\cos30^\circ=OL/OA\:and\:\cos60^\circ=OL/OB$
Now, let us substitute the corresponding values, we get
\[ \Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{p}{{OA}}{\rm{ and }}\dfrac{1}{2} = \dfrac{p}{{OB}}\]
On solving for\[OA\& OB\], we get
\[ \Rightarrow OA = \dfrac{{2p}}{{\sqrt 3 }}{\rm{ and }}OB = 2p\]
We have been already given that the area of triangle \[{\rm{OAB}}\] is
\[\dfrac{{50}}{{\sqrt 3 }}\]
By using triangle formula, we have to solve
\[\dfrac{1}{2} \times OA \times OB = \dfrac{{50}}{{\sqrt 3 }}\]
Now, let’s substitute the value, we get
\[ \Rightarrow \dfrac{1}{2} \times \dfrac{{2p}}{{\sqrt 3 }} \times 2p = \dfrac{{50}}{{\sqrt 3 }}\]
Now, we have to cancel similar terms, we get
\[ \Rightarrow {p^2} = 25\]
On squaring both sides, we obtain
\[ \Rightarrow p = 5\]
On substituting the value of \[p\] in (1), we get
\[\sqrt 3 x + y = 10\]
Therefore, the equation of the line \[AB\] is \[\sqrt 3 x + y \pm 10 = 0\].

Option ‘B’ is correct

Note: Three vital facts to recall in this question are: the locations where the line intersects the coordinate axes are called intercepts, the area of a right-angled triangle is \[\dfrac{1}{2} \times {\rm{height}} \times {\rm{base}}\] and the intercept version of the straight line equation is \[\dfrac{{\rm{x}}}{{\rm{a}}} + \dfrac{{\rm{y}}}{{\rm{b}}} = 1\].