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Distance formulae is the method to find the distance between two points S(x, y) and T(x, y) given by the formulae, \[D = \sqrt {{{\left( {{x_T} - {x_S}} \right)}^2} + {{\left( {{y_T} - {y_s}} \right)}^2}} \]

In the question, the locus of the point is desired such that the ratio of its distances from\[\left( {2,0} \right)\] and \[\left( {1,3} \right)\] is \[5:4\] for which consider a point on the line joining the points \[\left( {2,0} \right)\] and \[\left( {1,3} \right)\] and at the same time dividing the line segment in the ratio of \[5:4\] and use the distance formula accordingly.

Let us consider the point\[M(x,y)\]which lies between the point \[A\left( {2,0} \right)\]and\[B\left( {1,3} \right)\], the point \[M(x,y)\]divides the line into \[5:4\]ratio,

Now find the distance of point M from point A and from point B by using the distance formula

\[

{d_{MA}} = \sqrt {{{\left( {{x_M} - {x_A}} \right)}^2} + {{\left( {{y_M} - {y_A}} \right)}^2}} \\

= \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 0} \right)}^2}} \\

= \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( y \right)}^2}} \\

\]

And the distance from B

\[

{d_{MB}} = \sqrt {{{\left( {{x_M} - {x_B}} \right)}^2} + {{\left( {{y_M} - {y_B}} \right)}^2}} \\

= \sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 3} \right)}^2}} \\

\]

Given that the point \[M(x,y)\]divides the line AB into\[5:4\]ratio, hence we can write

\[\dfrac{{{d_{MA}}}}{{{d_{MB}}}} = \dfrac{5}{4}\]

This is equal to

\[

\dfrac{{{d_{MA}}}}{{{d_{MB}}}} = \dfrac{5}{4} \\

\dfrac{{\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( y \right)}^2}} }}{{\sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 3} \right)}^2}} }} = \dfrac{5}{4} \\

\dfrac{{{{\left( {x - 2} \right)}^2} + {y^2}}}{{{{\left( {x - 1} \right)}^2} + {{\left( {y - 3} \right)}^2}}} = \dfrac{{{5^2}}}{{{4^2}}} \\

\dfrac{{{{\left( {x - 2} \right)}^2} + {y^2}}}{{{{\left( {x - 1} \right)}^2} + {{\left( {y - 3} \right)}^2}}} = \dfrac{{25}}{{16}} \\

\]

Now use the cross multiplication to solve the equation further, hence we can write

\[

16\left\{ {{{\left( {x - 2} \right)}^2} + {y^2}} \right\} = 25\left\{ {{{\left( {x - 1} \right)}^2} + {{\left( {y - 3} \right)}^2}} \right\} \\

16\left\{ {{x^2} - 4x + 4 + {y^2}} \right\} = 25\left\{ {{x^2} - 2x + 1 + {y^2} - 6y + 9} \right\} \\

16{x^2} - 64x + 16{y^2} + 64 = 25{x^2} - 50x + 25 + 25{y^2} - 150y + 225 \\

9{x^2} + 9{y^2} + 14x - 150y + 186 = 0 \\

\]

Hence we get the equation of the locus at the point \[M(x,y)\]equal to \[9{x^2} + 9{y^2} + 14x - 150y + 186 = 0\]