Question

Let \(A(3,4)\), \(B(5,-2)\) and \(P(\alpha,\beta)\), \(\alpha\beta \neq 0\), be three points such that \(PA = PB\) and the area of \(\triangle PAB\) is \(10\). Then the distance of the point \(Q(2\alpha-5\beta,\; \alpha-\beta^2)\) from the line having intercepts \(3\) and \(1\) on the \(x\)- and \(y\)-axis respectively, is:

• A. \(10\)
• B. \(\sqrt{10}\)
• C. \(15\)
• D. \(2\)

Hint:

For problems involving equal distances from two points, always use the perpendicular bisector to simplify calculations.

Answer 1

The Correct Option is B

Concept:
A point equidistant from \(A\) and \(B\) lies on the perpendicular bisector of \(\overline{AB}\).
Area of a triangle \(=\dfrac{1}{2}\times\text{base}\times\text{height}\).
Distance of a point \((x_1,y_1)\) from a line \(ax+by+c=0\) is \[ \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}. \]
Step 1: Equation of the perpendicular bisector of \(AB\) Midpoint of \(AB\): \[ M\left(\frac{3+5}{2},\frac{4-2}{2}\right)=(4,1) \] Slope of \(AB\): \[ m_{AB}=\frac{-2-4}{5-3}=-3 \] Slope of perpendicular bisector \(=\frac{1}{3}\). Hence its equation is: \[ y-1=\frac{1}{3}(x-4) \]
Step 2: Use the area condition Length of \(AB\): \[ AB=\sqrt{(5-3)^2+(-2-4)^2}=\sqrt{40}=2\sqrt{10} \] Given area \(=10\), \[ 10=\frac{1}{2}\cdot 2\sqrt{10}\cdot h \Rightarrow h=\sqrt{10} \] Equation of line \(AB\) is: \[ y-4=-3(x-3)\Rightarrow 3x+y-13=0 \] Thus, \[ \frac{|3\alpha+\beta-13|}{\sqrt{10}}=\sqrt{10} \Rightarrow |3\alpha+\beta-13|=10 \]
Step 3: Find point \(P\) From perpendicular bisector, let \[ \alpha=4+3t,\quad \beta=1+t \] Substitute: \[ |3(4+3t)+(1+t)-13|=|10t|=10 \Rightarrow |t|=1 \] So \(t=1\) or \(-1\). Since \(\alpha\beta\neq0\), choose \(t=1\): \[ P=(7,2) \]
Step 4: Find point \(Q\) \[ Q(2\alpha-5\beta,\alpha-\beta^2)=(14-10,7-4)=(4,3) \]
Step 5: Distance of \(Q\) from the given line Line with intercepts \(3\) and \(1\): \[ \frac{x}{3}+\frac{y}{1}=1 \Rightarrow x+3y-3=0 \] Distance: \[ \frac{|4+9-3|}{\sqrt{1+9}}=\frac{10}{\sqrt{10}}=\sqrt{10} \]