Question

Find a point on the X-axis, which is equidistant from the points (7,6) and (3,4).​

Answer 1

Step-by-step explanation:

let suppose the ,,

7=x1

6=y1

and

3=x2

4=y2

Now see the attachments

Mark as brainlist plzzzzzz mate,☺️☺️☺️☺️☺️

Answer 2

⠀⠀ıllıllı uoᴉʇnloS ıllıllı

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

Let (a, 0) be the point of the x- axis that is

Equidistant from the points (7,6) and (3,4).

Accordingly

$\sqrt{(7 - a) {}^{2} + (6 - 0) ^{2} } = \sqrt{(3 - a) ^{2} + (4 - 0) ^{2} }$

$= \sqrt{49 + a ^{2} - 14a + 36} = \sqrt{9 + a^{2} - 6a + 16}$

$= \sqrt{a ^{2} - 14a + 85} = \sqrt{a ^{2} - 6a + 25}$

On squaring both sides, We obtain

${a}^{2} - 14a + 85 = a^{2} - 6a + 25$

$= - 14a + 6a = 25 - 85 \\ = - 8a = - 60 \\ = a = \frac{60}{8} = \frac{15}{2}$

Thus, the required points on the x-axis is :

• $( \frac{15}{2} , 0)$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$