Question

Find the point an x-axis which is equilateral from the point ( 7, 6 ) & ( -3, 4)

Answer 1

Given

A(7,6) B(-3,4)

To find

The point on x-axis equidistant from the points A and B

Explanation

As the point is present on x-axis ,

y-coordinate=0

let x coordinate=x

let the point be P

As it is equidistant from the points A(7, 6 ) & B( -3, 4)

so

$PA=PB$

$\boxed{\bf distance \: between \: two \:points=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Finding PA :(x,0) and (7,6)

$PA=\sqrt{(7+x)^2+(6-0)^2}$

$=>PA=\sqrt{ x^2+49+14x+36}=\sqrt{x^2+14x+85}units$

Finding PB: (x,0) and (-3,4)

$=>PB=\sqrt{(x+3)^2+4^2}$

$=>PB=\sqrt{x^2+9+6x+16}=\sqrt{x^2+6x+25}units$

now

$PA=PB$

Squaring on both sides

$PA^2=PB^2$

Substituting the values

$= > {x}^{2} + 14x + 85 = {x}^{2} + 6x + 25$

cancelling x²

$= > 14x + 85 = 6x + 25 \\ \\ = > 8x = - 60 \\ \\ \boxed{x = \dfrac{ - 15}{2} }$

So the point P (x,0)=(-15/2,0)

Answer 2

ANSWER:-

Given:

•Equidistant from the point (7,6) &(-3,4)

To find:

Find the point an x -axis.

Solution:

We have to find a point on x-axis.

So, its y-coordinate will be 0

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

Using Formula of the distance:

⚫A(x1,y1)

⚫B(x2,y2)

$AB = \sqrt{[(x2 - x1) {}^{2} + (y2 - y1) {}^{2} } ]$

Distance between: (x,0) & (7,6)

=) √[(x-7)² + (0-6)²]

=) √[(x-7)² + (-6)²].............(1)

&

Distance between: (x,0) & (-3,4)

=) √[(x+3)² +(0-4)²]

=) √[(x+3)² + (-4)²]..............(2)

Comparing equation, these distance are equal measure,

=)√[(x-7)² + (-6)²] = √[(x+3)² +(-4)²]

Squaring both sides, we get;

=) (x-7)² + 36 = (x+3)² + 16

Using [a+b]² & [a-b]²

=) x² +7² -2×x×7+36= x² +3² +2×x×3 +16

=) x² + 49 -14x +36 = x²+9 + 6x +16

=) x² + 85 -14x = x² + 25 +6x

=) -14x -6x = 25 -85

=) -20x = -60

=) x= -60/-20 [minus cancel]

=) x= 3

Thus,

The point is (3,0).

Thank you.