Question

Find the equation of the locus of a point which is always equidistant from the points whose coordinates are $(2,3)and(4,5)$.

Answer 1

Hint: We will consider a point P(h, k).Then we will use distance formula to calculate PA and PB. Thereafter, we will apply the equidistant condition to get the answer. By using the distance formula $ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $


Complete step by step solution:
Let the point be $P(h,k)$
Here, $A(2,3)$ and $B(4,5)$
Now, we use distance formula, for calculate the value of$PA\,\,and\,\,PB$
Distance formula$ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Now,$PA = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
$P(h,k)\,\,and\,\,A(2,3)$and$B(4,5)$ we substitute the value of $(x,y)and({x_2},{y_2})$
$PA = \sqrt {{{(h - 2)}^2} + {{(k - 3)}^2}} $
And $PB = \sqrt {{{(h - 4)}^2} + {{(k - 5)}^2}} $
Given $PA$and $PB$are equidistant then
$PA = PB$
\[\sqrt {{{(h - 2)}^2} + {{(k - 3)}^2}} = \sqrt {{{(h - 4)}^2} + {{(k - 5)}^2}} \]
Squaring both sides, we have
$ \Rightarrow {(h - 2)^2} + {(k - 3)^2} = {(h - 4)^2} + {(k - 5)^2}$
Now, we will using squaring the number using identity ${(a - b)^2} = {a^2} + {b^2} - 2ab$then
${x^2} + {(2)^2} - 2 \times 2h + {k^2}{(3)^2} - 2 \times 3 \times k = {(h)^2} + {(4)^2} - 2 \times 4 \times 1 + {(k)^2} + {(5)^2} - 2 \times k \times 5$
$ \Rightarrow {h^2} + 4 - 4h + {k^2} + 9 - 6k = {h^2} + 18 - 8h{k^2} + 25 - 10k$
$ \Rightarrow 4 - 4h + 9 - 6k = 16 - 8h + 25 - 10k$
$ \Rightarrow 4 + 4h + 9 - 6 - 16 + 8h - 35 + 10k = 0$
$ \Rightarrow 4 + 9 - 16 - 25 - 4h + 8h - 6k + 10k = 0$
$ \Rightarrow 15 - 41 + 4h + 4k = 0$
$ \Rightarrow - 28 + 4h + 4k = 0$
$ \Rightarrow 4( - 7 + h + k) = 0$
$ \Rightarrow - 7 + h + k = 0$
$ \Rightarrow h + k = 7$
$ \Rightarrow h + k - 7 = 0$
Therefore, the equation of line is $h + k - 7 = 0$


Note: Students must keep in mind that if a point P is equidistant from point A and other point B then it will the same distance from them .In other words ,PA=PB