Find a point on the X-axis, which is equidistant from the points (7,6)

1.	Find a point on the X-axis, which is equidistant from the points (7,6) and (3,4).
Answer» ⠀⠀ıllıllı uoᴉʇnloS ıllıllı $\setlength{\unitlength}{1.<klux>0</klux> cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$ 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 : $( \frac{15}{2} , 0)$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$