Question

A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of 60º with the line x + y = 0. Then an equation of the line L is:
(a) \[\left( \sqrt{3}+1 \right)x+\left( \sqrt{3}-1 \right)y=8\sqrt{2}\]
(b) \[\left( \sqrt{3}-1 \right)x+\left( \sqrt{3}+1 \right)y=8\sqrt{2}\]
(c) \[\sqrt{3}x+y=8\]
(d) \[x+\sqrt{3}y=8\]

Answer 1

Hint: Given angle between perpendicular and x + y = 0 as 60. Find the slope of lines by the given formula. From slopes by using the angle between lines formula find the condition on the required slope of line L. From the slope condition you get 2 slopes possible satisfying given conditions. Find the line equations with those slopes. From the line equations substitute x = 0 to get the y intercept and substitute y = 0 to get the x intercept. By using the condition that the line makes positive intercepts you can eliminate one of the lines and the other line will be the result.

Complete step by step answer:
If the angle between two lines with slopes a, b is X then
\[\tan X=\left| \dfrac{a-b}{1+ab} \right|\]
The slope of line ax + by + c = 0 is given by:
\[slope=-\dfrac{a}{b}\].
Let us assume the required line to be AB.
Let the origin be O.
The distance of line from origin is 4 units.
So OP = 4.

Let slope of OP be m.
Given that angle between OP and x + y = 0 is 60.
We need slope of x + y = 0.
Let us assume the slope to be k.
By using slope condition:
The slope of line ax + by + c = 0 is given by:
\[slope=-\dfrac{a}{b}\]
Using this condition on x + y = 0, we get:
a = 1
b = 1
By substituting values of a, b in condition, we get:
k = -1.
By using the angle between lines formula:
If the angle between two lines with slopes a, b is X then
\[\tan X=\left| \dfrac{a-b}{1+ab} \right|\].
X = 60.
a = m,
b = -1.
By substituting values of X, a, b in formula, we get:
\[\begin{align}
  & \tan 60=\left| \dfrac{m-1}{1+m} \right| \\
 & \dfrac{m-1}{1+m}=\pm \tan 60=\pm \sqrt{3} \\
\end{align}\]
Now we will take positive and negative cases separately and find the value of m in both cases.
Case-1
\[\dfrac{m-1}{m+1}=\sqrt{3}\]

\[m-1=\sqrt{3}m+\sqrt{3}\]
By simplifying, we get:
\[m=\dfrac{\sqrt{3}+1}{1-\sqrt{3}}\]
The line equation of a line with angle \[\alpha \]and perpendicular distance p is given by:
\[x\sin \alpha +y\cos \alpha =p.....(1)\]
Here in our case,
P = 4
\[\alpha ={{\tan }^{-1}}\dfrac{\sqrt{3}+1}{1-\sqrt{3}}\]
The value of sin\[\alpha \]:
\[\sin \alpha =\dfrac{\tan \alpha }{\sqrt{1+{{\tan }^{2}}\alpha }}=\dfrac{\left( \dfrac{\sqrt{3}+1}{1-\sqrt{3}} \right)}{\sqrt{1+{{\left( \dfrac{\sqrt{3}+1}{1-\sqrt{3}} \right)}^{2}}}}\]
From above, we get:
\[\sin \alpha =\dfrac{\sqrt{3}+1}{2\sqrt{2}}\]
The value of cos\[\alpha \]:
\[\cos \alpha =\dfrac{1}{\sqrt{1+{{\tan }^{2}}\alpha }}=\dfrac{1}{\sqrt{1+{{\left( \dfrac{\sqrt{3}+1}{1-\sqrt{3}} \right)}^{2}}}}\]
From above, we get:
\[\cos \alpha =\dfrac{\sqrt{3}-1}{2\sqrt{2}}\]
By finding sine and cosine of \[\alpha \], we get:
\[\sin \alpha =\dfrac{\sqrt{3}+1}{2\sqrt{2}}\text{ }and\text{ }\cos \alpha =\dfrac{\sqrt{3}-1}{2\sqrt{2}}\]
By substituting the values of sin \[\alpha \], cos \[\alpha \] and p in equation (1), we get:
\[\left( \dfrac{\sqrt{3}+1}{2\sqrt{2}} \right)x+\left( \dfrac{\sqrt{3}-1}{2\sqrt{2}} \right)y=4\]
By simplifying, we get:
\[\left( \sqrt{3}+1 \right)x+\left( \sqrt{3}-1 \right)y=8\sqrt{2}.....\left( 2 \right)\]
Case-2
\[\dfrac{m-1}{m+1}=-\sqrt{3}\]
\[m-1=-\sqrt{3}m-\sqrt{3}\]
By simplifying, we get:
\[m=\dfrac{-\sqrt{3}+1}{1+\sqrt{3}}\]
The line equation of a line with angle \[\alpha \] and perpendicular distance p is given by:
\[x\sin \alpha +y\cos \alpha =p.....(3)\]
Here in our case,
P = 4
\[\alpha ={{\tan }^{-1}}\dfrac{-\sqrt{3}+1}{1+\sqrt{3}}\]
The value of sin\[\alpha \]:
\[\sin \alpha =\dfrac{\tan \alpha }{\sqrt{1+{{\tan }^{2}}\alpha }}=\dfrac{\left( \dfrac{-\sqrt{3}+1}{1+\sqrt{3}} \right)}{\sqrt{1+{{\left( \dfrac{-\sqrt{3}+1}{1+\sqrt{3}} \right)}^{2}}}}\]
From above, we get:
\[\sin \alpha =\dfrac{-\sqrt{3}+1}{2\sqrt{2}}\]
The value of cos\[\alpha \]:
\[\cos \alpha =\dfrac{1}{\sqrt{1+{{\tan }^{2}}\alpha }}=\dfrac{1}{\sqrt{1+{{\left( \dfrac{-\sqrt{3}+1}{1+\sqrt{3}} \right)}^{2}}}}\]
From above, we get:
\[\cos \alpha =\dfrac{\sqrt{3}+1}{2\sqrt{2}}\]
By finding sine and cosine of \[\alpha \], we get:
\[\sin \alpha =\dfrac{-\sqrt{3}+1}{2\sqrt{2}}\text{ }and\text{ }\cos \alpha =\dfrac{\sqrt{3}+1}{2\sqrt{2}}\]
By substituting the values of sin \[\alpha \], cos \[\alpha \]and p in equation (3), we get:
\[\left( \dfrac{-\sqrt{3}+1}{2\sqrt{2}} \right)x+\left( \dfrac{\sqrt{3}+1}{2\sqrt{2}} \right)y=4\]
By simplifying, we get:
\[\left( -\sqrt{3}+1 \right)x+\left( \sqrt{3}+1 \right)y=8\sqrt{2}.....\left( 4 \right)\]
X and Y intercept of ax + by = c is given by:
\[x-\text{intercept}=\dfrac{c}{a}\text{ and y-intercept = }\dfrac{c}{b}\]
By applying this condition to equation (2), we get:
\[x-\text{intercept}=\dfrac{8\sqrt{2}}{\sqrt{3}+1}\text{ and y-intercept = }\dfrac{8\sqrt{2}}{\sqrt{3}-1}\]
By applying this condition to equation (4), we get:
\[x-\text{intercept}=\dfrac{8\sqrt{2}}{-\sqrt{3}+1}\text{ and y-intercept = }\dfrac{8\sqrt{2}}{\sqrt{3}+1}\]
You can see the x intercept in case of equation (4) is negative. So we are discarding that equation as in question its given line makes positive intercepts.
\[\therefore \]Equation (2) is the required line.
\[\left( \sqrt{3}+1 \right)x+\left( \sqrt{3}-1 \right)y=8\sqrt{2}\]

So, the correct answer is “Option A”.

Note: Be careful while applying tangent formula. Generally students take 1-ab as a denominator instead of 1+ab. This change of single sign may result in a great change in the whole result. Some students forget to take modulus thus end up getting only one equation. Which may be correct or not. So it is better to always consider the modulus symbol carefully.