Question

what point on x-axis is equidistant from the points A (7, 6) and B (-3, 4) ? ​

Answer 1

Answer:

(3,0)

Step-by-step explanation:

Let the required point on the x-axis be (x,0)

Given (x,0) is equidistant from (7,6) and (−3,4)

∴Distance between (x,0)≡(x1 ,y1 ) and (7,6)=(x2,y2)

distance between (x,0)≡(x 1 ,y 1) and (−3,4)≡(x2,y2 )

Distance Formula = (x 2−x 1 ) ^2+(y 2 −y 1 )^2

⇒ (7−x) 2 +(6−0) 2

= (−3−x) 2+(4−0) 2

⇒49−14x+x 2+36

=9+6x+x 2+16

⇒85−14x=25+6x

⇒−14x−6x=25−85

⇒−20x=−60

⇒x=3

∴ Required point on the x−axis=(3,0)

Answer 2

Answer :

(3,0)

Given :

• Point on x-axis is equidistant from the points A (7, 6) and B (-3, 4)

To find :

• Point on x-axis

Solution :

let the point on x-axis be (x,0)

• The point is equidistant from the points A (7, 6) and B (-3, 4)

• The distance between two points (a,b) and (p,q) is given by,

              $=\sqrt{(p-a)^2+(q-b)^2}$

• The distance between (7,6) and (x,0) = The distance between (x,0) and  (-3,4)

         $\sqrt{(7-x)^2+(6-0)^2} =\sqrt{(x-(-3))^2+(0-4)^2} \\\\ (7-x)^2+6^2=(x+3)^2+(-4)^2 \\\\ (7-x)^2+36=(x+3)^2+16 \\\\ 7^2+x^2-2(x)(7)+36=x^2+3^2+2(x)(3)+16 \\\\ 49+x^2-14x +36=x^2+9+6x+16 \\\\ x^2-14x+85 =x^2+6x+25 \\\\ -14x+85=6x+25 \\\\ 6x+14x=85-25 \\\\ 20x=60\\\\ x=60/20 \\\\ x=3$

the point on x-axis is (3,0)