Dear student, Ordinate of point A=xAbscissa of point A=2xThus, Coordinate of A is 2x,xCoordiante of B is 10,0Distance of AB=y2-y12+x2-x12⇒5=0-x2+10-2x2Squaring both sides⇒25=x2+100+4x2-40x --[a-b2=a2+b2-2ab]⇒25=5x2-40x+100⇒5x2-40x+100-25=0⇒5x2-40x+75=0⇒x2-8x+15=0⇒x2-5x-3x+15=0⇒xx-5-3x-5=0⇒x-5x-3=0⇒x=5 or x=3Thus the coordinate of A (2x,x)= 10,5 or 6,3 Regards

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Dear student,

O r d i n a t e o f p o i n t A = x A b s c i s s a o f p o i n t A = 2 x T h u s, C o o r d i n a t e o f A i s (2 x, x) C o o r d i a n t e o f B i s (10, 0) D i s \tan c e o f A B = \sqrt{{(y_{2} - y_{1})}^{2} + {(x_{2} - x_{1})}^{2}} \Rightarrow 5 = \sqrt{{(0 - x)}^{2} + {(10 - 2 x)}^{2}} S q u a r i n g b o t h s i d e s \Rightarrow 25 = x^{2} + 100 + 4 x^{2} - 40 x - - [{(a - b)}^{2} = a^{2} + b^{2} - 2 a b] \Rightarrow 25 = 5 x^{2} - 40 x + 100 \Rightarrow 5 x^{2} - 40 x + 100 - 25 = 0 \Rightarrow 5 x^{2} - 40 x + 75 = 0 \Rightarrow x^{2} - 8 x + 15 = 0 \Rightarrow x^{2} - 5 x - 3 x + 15 = 0 \Rightarrow x (x - 5) - 3 (x - 5) = 0 \Rightarrow (x - 5) (x - 3) = 0 \Rightarrow x = 5 o r x = 3 T h u s t h e c o o r d i n a t e o f A (2 x, x) = (10, 5) o r (6, 3)

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