What is the equation to the straight line passing through the point $(-sin\theta, cos\theta)$ and perpendicular to the line $xcos\theta + ysin\theta = 9$?
$xsin\theta - ycos\theta - 1 = 0$
$xsin\theta - ycos\theta + 1 = 0$
$xsin\theta - ycos\theta = 0$
$xcos\theta - ysin\theta + 1 = 0$
Two points $P$ and $Q$ lie on line $y = 2x + 3$. These two points $P$ and $Q$ are at a distance 2 units from another point $R(1, 5)$. What are the coordinates of the points $P$ and $Q$?
$(1 + \frac{2}{\sqrt{5}}, 5 + \frac{4}{\sqrt{5}}), (1 - \frac{2}{\sqrt{5}}, 5 - \frac{4}{\sqrt{5}})$
$(3 + \frac{2}{\sqrt{5}}, 5 + \frac{4}{\sqrt{5}}), (-1 - \frac{2}{\sqrt{5}}, 5 - \frac{4}{\sqrt{5}})$
$(1 - \frac{2}{\sqrt{5}}, 5 + \frac{4}{\sqrt{5}}), (1 + \frac{2}{\sqrt{5}}, 5 - \frac{4}{\sqrt{5}})$
$(3 - \frac{2}{\sqrt{5}}, 5 + \frac{4}{\sqrt{5}}), (-1 + \frac{2}{\sqrt{5}}, 5 - \frac{4}{\sqrt{5}})$
If two sides of a square lie on the lines $2x + y - 3 = 0$ and $4x + 2y + 5 = 0$, then what is the area of the square in square units?
6.05
6.15
6.25
6.35
ABC is a triangle with A(3, 5). The mid-points of sides AB, AC are at (-1, 2), (6, 4) respectively. What are the coordinates of centroid of the triangle ABC?
$\left(\frac{8}{3}, \frac{11}{3}\right)$
$\left(\frac{7}{3}, \frac{7}{3}\right)$
$\left(2, \frac{8}{3}\right)$
$\left(\frac{8}{3}, 2\right)$