The equations of the lines on which the perpendicular from the origin make 30 angle with x-axis and which form a triangle of area 50/3 with axes, are

Question

The equations of the lines on which the perpendicular from the origin make $30 °$ angle with $x$ -axis and which form a triangle of area $\frac{50}{\sqrt{3}}$ with axes, are

Smart Achievers · Accepted Answer

(2)

$Let p be the length of perpendicular from the origin$ $on the given line. Then its equation in normal form is given by$

$x \cos 30 ° + y \sin 30 ° = p or \sqrt{3} x + y = 2 p$

$This meets with the coordinate axes at$ $A (\frac{2 p}{\sqrt{3}}, 0) and B (0, 2 p) .$

$∴$ $Area of ∆ O A B = \frac{1}{2} (\frac{2 p}{\sqrt{3}}) (2 p) = \frac{2 p^{2}}{\sqrt{3}}$

$Also, it is given that$ $\frac{2 p^{2}}{\sqrt{3}} = \frac{50}{\sqrt{3}} \Rightarrow p = \pm 5$

$Hence, the required equations of the lines are$ $\sqrt{3} x + y \pm 10 = 0$

Solution

Q.
The equations of the lines on which the perpendicular from the origin make $30 °$ angle with $x$ -axis and which form a triangle of area $\frac{50}{\sqrt{3}}$ with axes, are

1 $x + \sqrt{3} y \pm 10 = 0$

2 $\sqrt{3} x + y \pm 10 = 0$

3 $x \pm \sqrt{3} y - 10 = 0$

4 None of the above

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Solution

Q. The equations of the lines on which the perpendicular from the origin make 30° angle with x-axis and which form a triangle of area 503 with axes, are

1 x+3y±10=0

2 3x+y±10=0