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).