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Question: The point P is equidistant from \(A\left( 1,3 \right),B\left( -3,5 \right)\) and \(C\left( 5,-1 \rig...

The point P is equidistant from A(1,3),B(−3,5)A\left( 1,3 \right),B\left( -3,5 \right) and C(5,−1)C\left( 5,-1 \right), then PAPA is equal to:
A. 5
B. 555\sqrt{5}
C. 25
D. 5105\sqrt{10}

Explanation

Solution

In order to solve the problem, we need to find the distance from the first point to a certain point and from the second point to the same point. And as the distance between them is the same, then we can equate them by using the distance formula. The distance formula says that Distance = (x2−x1)2+(y2−y1)2\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}, where x2,x1{{x}_{2}},{{x}_{1}} are the x-coordinates and y2,y1{{y}_{2}},{{y}_{1}} are the y-coordinates.

Complete step-by-step solution:
We need to find the point P which is equidistant to point A and B and C. Let the point A be A(1,3)A\left( 1,3 \right), let the point B be B(−3,5)B\left( -3,5 \right) and let point C be C(5,−1)C\left( 5,-1 \right).
Let P=(x,y)P=\left( x,y \right) , then PA=PBPA=PB and PB=PCPB=PC.
∴PA2=PB2 ⇒(x−1)2+(y−3)2=(x+3)2+(y−5)2 ⇒x2−2x+1+y2−6y+9=x2+6x+9+y2−10y+25 ⇒−8x+4y−24=0 ⇒2x−y+6=0………(i) \begin{aligned} & \therefore P{{A}^{2}}=P{{B}^{2}} \\\ & \Rightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{\left( x+3 \right)}^{2}}+{{\left( y-5 \right)}^{2}} \\\ & \Rightarrow {{x}^{2}}-2x+1+{{y}^{2}}-6y+9={{x}^{2}}+6x+9+{{y}^{2}}-10y+25 \\\ & \Rightarrow -8x+4y-24=0 \\\ & \Rightarrow 2x-y+6=0\ldots \ldots \ldots \left( i \right) \\\ \end{aligned}
Now it is also clear that PB=PCPB=PC. So,
PB2=PC2 ⇒(x+3)2+(y−5)2=(x−5)2+(y+1)2 ⇒x2+6x+9+y2−10y+25=x2−10x+25+y2+2y+1 ⇒16x−12y+8=0 \begin{aligned} & P{{B}^{2}}=P{{C}^{2}} \\\ & \Rightarrow {{\left( x+3 \right)}^{2}}+{{\left( y-5 \right)}^{2}}={{\left( x-5 \right)}^{2}}+{{\left( y+1 \right)}^{2}} \\\ & \Rightarrow {{x}^{2}}+6x+9+{{y}^{2}}-10y+25={{x}^{2}}-10x+25+{{y}^{2}}+2y+1 \\\ & \Rightarrow 16x-12y+8=0 \\\ \end{aligned}
And it can be further written as follows,
⇒4x−3y+2=0………(ii)\Rightarrow 4x-3y+2=0\ldots \ldots \ldots \left( ii \right)
From equation (i) we get, y=2x+6y=2x+6, so putting this in equation (ii), we will get as follows,
4x−3(2x+6)+2=0 ⇒4x−6x−18+2=0 ⇒−2x−16=0 ⇒x=−8 \begin{aligned} & 4x-3\left( 2x+6 \right)+2=0 \\\ & \Rightarrow 4x-6x-18+2=0 \\\ & \Rightarrow -2x-16=0 \\\ & \Rightarrow x=-8 \\\ \end{aligned}
Therefore we get the value of y=2×(−8)+6=−10y=2\times \left( -8 \right)+6=-10.
So, the value of PAPA will be,
PA=(−8+3)2+(−10−5)2 =(−5)2+(−15)2 =250=510 \begin{aligned} & PA=\sqrt{{{\left( -8+3 \right)}^{2}}+{{\left( -10-5 \right)}^{2}}} \\\ & =\sqrt{{{\left( -5 \right)}^{2}}+{{\left( -15 \right)}^{2}}} \\\ & =\sqrt{250}=5\sqrt{10} \\\ \end{aligned}
Hence the correct answer is option D.

Note: A point is said to be equidistant if the distance between that point to the given point is found to be equal. For example, in the circle, the centre of the circle is equidistant from every point on the circle. Similarly in a rectangle, the centre of the rectangle is equidistant from all the four vertices.