Explanation
Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and$$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using theformula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{2 }-{ y }_{ 1 } \right) ^{ 2 } } $$Distance between the points $$ A(4,1) $$ and $$ B (3,a) = \sqrt {10} $$
$$ => \sqrt { \left( 3-4 \right) ^{ 2 }+\left( a-1 \right) ^{ 2 } } = \sqrt {10} $$
$$ \sqrt { 1 + {(a-1) }^{ 2 }} = \sqrt {10} $$
$$ 1 + {(a-1) }^{ 2 } = 10 $$
$$ { (a-1) }^{ 2 } = 9 $$
$$ a-1 = \pm 3 $$
$$ => a-1 = 3 ; a-1 = -3 $$
Let the point on the x-axis be $$ (x,0) $$ Distance between $$ (x,0) $$ and $$ (7,6) = \sqrt { \left( 7-x \right) ^{ 2}+\left( 6 - 0 \right) ^{ 2 } } = \sqrt { { 7}^{ 2 }+ { x }^{ 2 } - 14x + 36 }= \sqrt { { x }^{ 2 } - 14x + 85 } $$Distance between $$ (x,0) $$ and $$ (-3,4) = \sqrt { \left( -3-x \right) ^{ 2}+\left( 4 - 0 \right) ^{ 2 } } = \sqrt { { 3}^{ 2 }+ { x }^{ 2 } + 6x + 16 } =\sqrt { { x }^{ 2 } + 6x + 25 } $$As the point $$ (x,0) $$ is equidistant from the two points, both the distancescalculated are equal. $$ \sqrt { { x }^{ 2 } - 14x + 85 } = \sqrt { { x }^{ 2 } + 6x + 25 } $$$$ => { x }^{ 2 } - 14x + 85 = { x }^{ 2 } + 6x + 25 $$$$ 85 - 25 = 6x + 14x $$$$ 60 = 20x $$$$ x = 3 $$Thus, the point is $$ (3,0) $$
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