Straight Lines
Straight Line
Allen Star Batch
Grade 11
Question:
$A(1,2)$ and $B(7,10)$ are two points. If $P(x,y)$ is a point such that the angle $APB$ is $60°$ and the area of $\triangle APB$ is maximum, then which of given is (are) true?
$P$ lies on any line perpendicular to $AB$
$P$ lies on the right bisector of $AB$
$P$ lies on the straight line $3x + 4y = 36$
$P$ lies on the circle passing through the points $(1,2)$ and $(7,10)$ and having a radius of $10$ units
Step-by-Step Solution
Key Concept: Maximum area occurs when triangle $APB$ is equilateral, making the altitude and median coincide.
The area of triangle $APB$ is $\frac{1}{2} \times AP \times PB \times \sin(60°) = \frac{1}{2}AB \times h$. Using the constraint that $\angle APB = 60°$ and simplifying with the distance formula, we get $h = \frac{20}{\sqrt{3}}\sin\theta\sin(120-\theta) - \frac{10}{\sqrt{3}}[\cos(2\theta-120)-\cos(120)]$. For maximum area, we need $2\theta - 120 = \theta$, giving $\theta = 60°$, making the triangle equilateral. The perpendicular $PD$ acts as both bisector and median, with equation $y - 6 = -\frac{6}{8}(x-4)$, which simplifies to $3x + 4y = 36$. Point $P$ lies on the right bisector of $AB$.
Correct Answer: 2,3