Question

If a line L passing through the point A(-2,4) makes an angle of 60$^\circ$ with the positive direction of X-axis in anti-clockwise direction and B(p,q) lying in the 3$^\text{rd}$ quadrant is a point on L at the distance of 6 units from the point A, then $\sqrt{p^2+q^2-8q} =$

• A. 6
• B. 7
• C. 8
• D. 9

Hint:

The parametric form of a line is extremely useful for finding points at a specific distance from a given point along a certain direction. Remember that the distance 'r' is a directed distance, so it can be positive or negative depending on the direction from the initial point.

Answer 1

The Correct Option is A

We use the parametric form of a line to find the coordinates of point B.
The parametric equations of a line passing through $(x_1, y_1)$ at an angle $\theta$ are:
$x = x_1 + r\cos\theta$ and $y = y_1 + r\sin\theta$, where $r$ is the directed distance.
Here, $(x_1, y_1) = A(-2, 4)$, $\theta = 60^\circ$, and the distance is 6 units. So, $r = \pm 6$.
The coordinates of B are $(p, q)$.
$p = -2 + r\cos(60^\circ) = -2 + r(\frac{1}{2})$.
$q = 4 + r\sin(60^\circ) = 4 + r(\frac{\sqrt{3}}{2})$.
Since B(p,q) lies in the 3rd quadrant, both $p$ and $q$ must be negative.
$p<0 \implies -2 + \frac{r}{2}<0 \implies \frac{r}{2}<2 \implies r<4$.
$q<0 \implies 4 + \frac{r\sqrt{3}}{2}<0 \implies \frac{r\sqrt{3}}{2}<-4 \implies r<-\frac{8}{\sqrt{3}}$.
Since $r$ must satisfy both conditions, it must be negative. We take the directed distance $r = -6$.
Now we calculate $p$ and $q$:
$p = -2 + (-6)(\frac{1}{2}) = -2 - 3 = -5$.
$q = 4 + (-6)(\frac{\sqrt{3}}{2}) = 4 - 3\sqrt{3}$.
Now we evaluate the expression $\sqrt{p^2+q^2-8q}$.
$p^2 = (-5)^2 = 25$.
$q^2 = (4-3\sqrt{3})^2 = 4^2 - 2(4)(3\sqrt{3}) + (3\sqrt{3})^2 = 16 - 24\sqrt{3} + 27 = 43 - 24\sqrt{3}$.
$-8q = -8(4-3\sqrt{3}) = -32 + 24\sqrt{3}$.
$p^2+q^2-8q = 25 + (43 - 24\sqrt{3}) + (-32 + 24\sqrt{3}) = 25 + 43 - 32 = 36$.
$\sqrt{p^2+q^2-8q} = \sqrt{36} = 6$.