Question

$A(1,2)$ and $B(7,10)$ are two points. If $p\left(x\right)$ is a point such that the $\angle APB$ is ${60}^{\circ }$ and the area of the $△APB$ is maximum, then which of the following is (are) true?
(A) $p$ lies on the straight line $3x+4y=36$
(B) $p$ lies on any line perpendicular to $AB$

Answer 1

In $△APB,\angle P={60}^{\circ }$.

$A(1,2),B(7,10),P(x,y)$

Area of $△APB$ will be maximum if it is an equilateral triangle.

Hence, the perpendicular bisector of $AB$ will pass through the point $P.$

Mid point of $AB=(\frac{1+7}{2},\frac{2+10}{2})=(4,6)$

Slope of $AB=\left(\frac{10-2}{7-1}\right)=\frac{4}{3}$

Slope of line $\perp$ to $AB=-\frac{3}{4}$

Therefore the equation of the straight line is:

$(y-y_1)=m(x-x_1)$

$\Rightarrow (y-6)=\frac{-3}{4}(x-4)$

$\Rightarrow 4y-24=-3x+12$

$\Rightarrow 3x+4y=36$

Hence $p$ lies on the straight line $3x+4y=36$ is true.