A is a point on x-axis, B = (5, 4) andAB = 5 units, find the co-ordinates of A. About the author Peyton

Answer: (8 , 0) (or) (2 , 0) Step-by-step explanation: Given, B = (5 , 4) As ‘A’ is the point on ‘x’ axis the co-ordinates of ‘A’ will be “(x , 0)” AB = 5units To Find :- Co-ordinates of point ‘A’ Formula Required :- [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex] Squaring on both sides :- [tex]\implies d^2=(x_2-x_1)^2+(y_2-y_1)^2[/tex] Solution :- [tex]x_1=x,y_1=0[/tex] [tex]x_2=5,y_2=4[/tex] AB = 5 Squaring on both sides :- [tex](AB)^2=(5)^2[/tex] [tex](5-x)^2+(4-0)^2=25[/tex] [tex]5^2-10x+x^2+4^2=25[/tex] [tex]x^2-10x+25+16-25=0[/tex] [tex]x^2-10x+16=0[/tex] [tex]x^2-8x-2x+16=0[/tex] [tex]x(x-8)-2(x-8)=0[/tex] [tex](x-8)(x-2)=0[/tex] x = 8 , 2 Co-ordinates of A = (x , 0) = (8 , 0) (or) (2 , 0) Reply

Answer:(8 , 0) (or) (2 , 0)

Step-by-step explanation:Given,B = (5 , 4)

As ‘A’ is the point on ‘x’ axis the co-ordinates of ‘A’ will be “(x , 0)”

AB = 5units

To Find :-Co-ordinates of point ‘A’

Formula Required :-[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

Squaring on both sides :-[tex]\implies d^2=(x_2-x_1)^2+(y_2-y_1)^2[/tex]

Solution :-[tex]x_1=x,y_1=0[/tex]

[tex]x_2=5,y_2=4[/tex]

AB = 5

Squaring on both sides :-

[tex](AB)^2=(5)^2[/tex]

[tex](5-x)^2+(4-0)^2=25[/tex]

[tex]5^2-10x+x^2+4^2=25[/tex]

[tex]x^2-10x+25+16-25=0[/tex]

[tex]x^2-10x+16=0[/tex]

[tex]x^2-8x-2x+16=0[/tex]

[tex]x(x-8)-2(x-8)=0[/tex]

[tex](x-8)(x-2)=0[/tex]

x = 8 , 2

Co-ordinates of A = (x , 0) = (8 , 0) (or) (2 , 0)