Question

If the distance from P to the points (2, 3), (2,-3) are in the ratio 2:3, then find the locus of P

Answer 1

Hint : We can solve this problem by using the distance between two points formula. If $A({x_1},{y_1})$ and $B({x_2},{y_2})$ are two points then the distance is given by $AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ . By using the given ratio and distance formula we can find the locus of P. A locus is a set of all the points whose position is defined by a certain condition.

Complete step-by-step answer :
Let, $A(2,3)$ and $B(2, - 3)$ be the two points.
$P(x,y)$ be the locus of the points of intersection.
The distance between $P(x,y)$ and $A(2,3)$ is $AP = \sqrt {{{(2 - x)}^2} + {{(3 - y)}^2}}$ .
The distance between $P(x,y)$ and $B(2, - 3)$ is $BP = \sqrt {{{(2 - x)}^2} + {{( - 3 - y)}^2}}$ .
Given, the distance from P to the points (2, 3), (2,-3) are in the ratio 2:3.
That is, $\dfrac{{AP}}{{BP}} = \dfrac{2}{3}$
Squaring on both sides, we get.
$\Rightarrow { \left( { \dfrac{{AP}}{{BP}}} \right)^2} = { \left( { \dfrac{2}{3}} \right)^2}$
Substituting, $AP$ and $BP$ in above equation,
We get, $\Rightarrow { \left( { \dfrac{{ \sqrt {{{(2 - x)}^2} + {{(3 - y)}^2}} }}{{ \sqrt {{{(2 - x)}^2} + {{( - 3 - y)}^2}} }}} \right)^2} = \dfrac{4}{9}$
Square root and square will get cancelled. We get,
$\Rightarrow \dfrac{{{{(2 - x)}^2} + {{(3 - y)}^2}}}{{{{(2 - x)}^2} + {{( - 3 - y)}^2}}} = \dfrac{4}{9}$
We know that, ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and ${(a - b)^2} = {a^2} + {b^2} - 2ab$ .
Applying this we get,
$\Rightarrow \dfrac{{(4 + {x^2} - 4x) + (9 + {y^2} - 6y)}}{{(4 + {x^2} - 4x) + (9 + {y^2} - 6y)}} = \dfrac{4}{9}$
We need to get this in an equation form so, cross multiplying, we get:
$\Rightarrow 9(4 + {x^2} - 4x + 9 + {y^2} - 6y) = 4(4 + {x^2} - 4x + 9 + {y^2} - 6y)$
Expanding the brackets,
$\Rightarrow 36 + 9{x^2} - 36x + 81 + 9{y^2} - 54y = 16 + 4{x^2} - 16x + 36 + 4{y^2} + 24y$
Rearranging the equation to one side,
$\Rightarrow 9{x^2} - 4{x^2} + 9{y^2} - 4{y^2} - 54y - 24y + 36 + 81 - 16 - 36 = 0$
Adding like terms and constant terms,
$\Rightarrow 5{x^2} + 5{y^2} - 20x - 78y + 65 = 0$
Hence the locus of P is $5{x^2} + 5{y^2} - 20x - 78y + 65 = 0$ .
So, the correct answer is “$5{x^2} + 5{y^2} - 20x - 78y + 65 = 0$”.

Note : Students need to remember the distance between two point formulas. Here the calculation part may look difficult but all we did is simple addition and multiplication. Also, while rearranging the equation do not miss the terms. They ask the same question with different points or different ratios. Use the same method which we have done above.