Question

Find the equation of the locus of a point $P$ which is equidistant from the straight line $3x - 4y + 2 = 0$ and the origin

Answer 1

Hint : The general coordinate of origin is $O(0,0)$ and let the given point has coordinate $P(x,y)$ .We need to equate the distance of the point $P$ from the origin $O$ and the straight line $3x - 4y + 2 = 0$ .After equating we will generate a new equation which will be the solution to the given problem.
Distance between any two points $A({x_1},{y_1})$ and $B({x_2},{y_2})$ is given by $d(A,B) = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}}$ .
The distance between any line $L:ax + by + c = 0$ and a point $A({x_1},{y_1})$ is given by $d(L,A) = \dfrac{{|a{x_1} + b{y_1} + c|}}{{\sqrt {{a^2} + {b^2}} }}$

Complete step-by-step answer :
We are given that $P(x,y)$ is equidistant from origin and the straight line $3x - 4y + 2 = 0$ .
The distance between Origin $O(0,0)$ and the point $P(x,y)$ is:
$d(O,P) = \sqrt {{{(0 - x)}^2} + {{(0 - y)}^2}}$
Since $0$ subtracted from any number gives us the same result we can simplify the above equation as:
$d(O,P) = \sqrt {{{(x)}^2} + {{(y)}^2}}$ ------(1)
Again the distance between the line $L:3x - 4y + 2 = 0$ and the point $P(x,y)$ is:
$d(L,P) = \dfrac{{|3x - 4y + 2|}}{{\sqrt {{3^2} + {{( - 4)}^2}} }}$ ------(2)
According to the question:
$d(L,P) = d(O,P)$
On substituting the values from (1) and (2) we get-
$\dfrac{{|3x - 4y + 2|}}{{\sqrt {{3^2} + {{( - 4)}^2}} }} = \sqrt {{{(x)}^2} + {{(y)}^2}}$
Squaring the above equation on both sides:
$\dfrac{{|3x - 4y + 2{|^2}}}{{{{(\sqrt {{3^2} + {{( - 4)}^2}} )}^2}}} = {(\sqrt {{{(x)}^2} + {{(y)}^2}} )^2}$
$\Rightarrow \dfrac{{|3x - 4y + 2{|^2}}}{{{3^2} + {{( - 4)}^2}}} = {x^2} + {y^2}$
On cross multiplication we get:
$|3x - 4y + 2{|^2} = ({x^2} + {y^2})({3^2} + {( - 4)^2})$
$\Rightarrow {(3x - 4y + 2)^2} = ({x^2} + {y^2})({3^2} + {( - 4)^2})$ [ $\because |a{|^2} = {a^2}$ ]
$\Rightarrow (3x - 4y + 2) \times (3x - 4y + 2) = ({x^2} + {y^2})(9 + 16)$
$\Rightarrow (3x - 4y + 2) \times (3x - 4y + 2) = ({x^2} + {y^2})(25)$
$\Rightarrow 3x(3x - 4y + 2) - 4y(3x - 4y + 2) + 2(3x - 4y + 2) = ({x^2} + {y^2})(25)$
$\Rightarrow (9{x^2} - 12xy + 6x) + ( - 12xy + 16{y^2} - 8y) + (6x - 8y + 4) = ({x^2} + {y^2})(25)$
$\Rightarrow 9{x^2} + 16{y^2} - 12xy - 12xy + 6x + 6x - 8y - 8y + 4 = ({x^2} + {y^2})(25)$
$\Rightarrow 9{x^2} + 16{y^2} - 24xy + 12x - 16y + 4 = 25{x^2} + 25{y^2}$
$\Rightarrow 25{x^2} + 25{y^2} - (9{x^2} + 16{y^2} - 24xy + 12x - 16y + 4) = 0$
$\Rightarrow 25{x^2} - 9{x^2} + 25{y^2} - 16{y^2} + 24xy - 12x + 16y - 4 = 0$
$\Rightarrow 16{x^2} + 9{y^2} + 24xy - 12x + 16y - 4 = 0$
Therefore, the locus of the point $P$ which is $16{x^2} + 9{y^2} + 24xy - 12x + 16y - 4 = 0$

Note : Here after substituting all the values we got a linear equation in 2 variables. Simplify the equation by adding coefficients of like terms one by one to avoid errors in calculating which may lead to a different equation. We take $P(x,y)$ because the locus is set of all points that satisfy a condition equidistant from origin and the line in this question).