Question

Show that the path of a moving point which remains at equal distance from the points \[\left( {2,1} \right)\] and $\left( { - 3, - 2} \right)$ is a straight line .

Answer 1

Hint: First let us suppose that the an arbitrary point $\left( {x,y} \right)$ , hence it is given that it is equidistant mean that AB = AC hence it mean $\sqrt {{{(x + 3)}^2} + {{(y + 2)}^2}} $ $ = $ $\sqrt {{{(x - 2)}^2} + {{(y - 1)}^2}} $ solve this equation and at last we get equation of line .

Complete step-by-step answer:
In this question we have show that the path of a moving point which remains at equal distance from the points \[\left( {2,1} \right)\] and $\left( { - 3, - 2} \right)$ is a straight line basically we have to find the locus of the point which is equidistant with \[\left( {2,1} \right)\] and $\left( { - 3, - 2} \right)$ ,
So for this let us suppose that the an arbitrary point $\left( {x,y} \right)$ which is equidistant with the points \[\left( {2,1} \right)\] and $\left( { - 3, - 2} \right)$ ,

So from the given question we know that
AB = AC ...........(i)
And the distance formula we know that the , $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
So distance AB = $\sqrt {{{(x - 2)}^2} + {{(y - 1)}^2}} $
Similarly for AC = $\sqrt {{{(x + 3)}^2} + {{(y + 2)}^2}} $
So from (i)
$\Rightarrow$ $\sqrt {{{(x + 3)}^2} + {{(y + 2)}^2}} $ $ = $ $\sqrt {{{(x - 2)}^2} + {{(y - 1)}^2}} $
On squaring both side we get ,
$\Rightarrow$ ${(x + 3)^2} + {(y + 2)^2}$ $ = {(x - 2)^2} + {(y - 1)^2}$
Now opening the brackets we get ,
$\Rightarrow$ ${x^2} + 6x + 9 + {y^2} + 4y + 4 = {x^2} - 4x + 4 + {y^2} - 2y + 1$
On cancelling the common terms we get ,
$\Rightarrow$ $6x + 4y + 13 = - 4x + 5 - 2y$
$\Rightarrow$ $6y = - 10x - 8$
$\Rightarrow$ $y = \dfrac{{ - 5}}{3}x - \dfrac{4}{3}$
Hence it is equation of the line that is in the form $y = mx + c$
where slope is m = $\dfrac{{ - 5}}{3}$ and c = $\dfrac{{ - 4}}{3}$
Proved .

Note: For finding the locus of the point first suppose any arbitrary point $\left( {x,y} \right)$ and then perform according to the question .
 The section formula $x = \dfrac{{m{x_1} + n{x_2}}}{{m + n}}$ or $y = \dfrac{{m{y_1} + n{y_2}}}{{m + n}}$ if we put $m = 1,n = 1$ then we can find the midpoint of the line using this .
If the product of the slope of the two lines is equal to the $ - 1$ then it would be perpendicular to each other .