Question

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

Answer 1

Let ( a, 0 ) be the point on 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 point on the x-axis is ( $\frac{15}{2}$,0)

Answer 2

Step-by-step explanation:

Let the point on the x-axis be (x,0)

Distance between (x,0) and (7,6)=(7−x)2+(6−0)2=72+x2−14x+36=x2−14x+85

Distance between (x,0) and (−3,4)=(−3−x)2+(4−0)2=32+x2+6x+16=x2+6x+25

As the point (x,0) is equidistant from the two points, both the distances

calculated are equal.

x2−14x+85=x2+6x+25

=>x2−14x+85=x2+6x+25

85−25=6x+14x

60=20x

x=3