Question

The
x-cordinate of a point P is twice
its y-cordinate. If P is equidistant from
Q (2,5) & R (-3, 6), then find the cordinate
of P​

Answer 1

Solution :-

Let the coordinates of point P be ( x , y ).

The point P is equidistant from Q ( 2 , 5 ) and R ( -3 , 6 ).

That is,

QP = RP

According to the first condition :-

$\longrightarrow \sf x = 2y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..................(1)$

According to the second condition :-

$\bullet \bf \ Distance \ formula = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\longrightarrow \sf QP = RP \\\\\longrightarrow \sqrt{(x-2)^2+(y+5)^2} = \sqrt{(x+3)^2+(y-6)^2} \\\\\longrightarrow (x-2)^2+(y+5)^2 = (x+3)^2+(y-6)^2\\\\\longrightarrow x^2+4-4x+y^2+25+10y = x^2+9+6x+y^2+36-12y \\\\\longrightarrow -4x+10y+29 = 6x-12y+45 \\\\\longrightarrow -4x-6x+10y+12y = 45-29 \\\\\longrightarrow -10x+22y = 16 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ....................(2)$

Substitute the value of x in eq(2) :-

$\longrightarrow \sf -2 y\times 10+22y = 16 \\\\\longrightarrow -20y+22y = 16 \\\\\longrightarrow 2y = 16 \\\\\longrightarrow \bf y = 8$

y coordinate = 8

x coordinate = 8 × 2 = 16

Coordinates of P = ( 16 , 8 )