Question

Find the locus of the points which are at equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$ and which are near the origin.
A. $21x - 77y + 153 = 0$
B. $99x + 77y - 133 = 0$
C. $7x - 11y = 19$
D. None of these

Answer 1

Hint: We will assume that the point $P\left( {h,k} \right)$ is at an equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$. By using the formula $d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$, we will calculate the distance of the point $P\left( {h,k} \right)$ from the given lines. Then equate the distances and replace $h$ by $x$ and $k$ by $y$. We will get two equations. Again, we find the distance of the lines from the origin and compare the distance. The shortest distance line is the required answer.

Formula Used:
The distance from a point $\left( {{x_1},{y_1}} \right)$ to a line $ax + by + c = 0$ is $d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$.

Complete step by step solution:
Let $P\left( {h,k} \right)$ be a point which is at an equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$.
Applying the formula $d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$ to calculate the distance between $P\left( {h,k} \right)$ and $3x + 4y - 11 = 0$.
$d = \left| {\dfrac{{3h + 4k - 11}}{{\sqrt {{3^2} + {4^2}} }}} \right|$
  $ = \left| {\dfrac{{3h + 4k - 11}}{{\sqrt {9 + 16} }}} \right|$
 $ = \left| {\dfrac{{3h + 4k - 11}}{5}} \right|$
Applying the formula $d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$ to calculate the distance between $P\left( {h,k} \right)$ and $12x + 5y + 2 = 0$.
$d = \left| {\dfrac{{12h + 5k + 2}}{{\sqrt {{{12}^2} + {5^2}} }}} \right|$
$ = \left| {\dfrac{{12h + 5k + 2}}{{\sqrt {144 + 25} }}} \right|$
$ = \left| {\dfrac{{12h + 5k + 2}}{{13}}} \right|$
Now we will equate the distances
$\left| {\dfrac{{3h + 4k - 11}}{5}} \right| = \left| {\dfrac{{12h + 5k + 2}}{{13}}} \right|$
Removing the modulus
$ \Rightarrow \dfrac{{3h + 4k - 11}}{5} = \pm \dfrac{{12h + 5k + 2}}{{13}}$
Now we will be replacing $h$ by $x$ and $k$ by $y$.
$ \Rightarrow \dfrac{{3x + 4y - 11}}{5} = \pm \dfrac{{12x + 5y + 2}}{{13}}$
Either or,
$ \Rightarrow \dfrac{{3x + 4y - 11}}{5} = \dfrac{{12x + 5y + 2}}{{13}}$ $ \Rightarrow \dfrac{{3x + 4y - 11}}{5} = - \dfrac{{12x + 5y + 2}}{{13}}$
Applying the cross multiplication
$ \Rightarrow 13\left( {3x + 4y - 11} \right) = 5\left( {12x + 5y + 2} \right)$ $ \Rightarrow 13\left( {3x + 4y - 11} \right) = - 5\left( {12x + 5y + 2} \right)$
$ \Rightarrow 39x + 52y - 143 = 60x + 25y + 10$ $ \Rightarrow 39x + 52y - 143 = - 60x - 25y - 10$
$ \Rightarrow 21x - 27y + 153 = 0$ $ \Rightarrow 99x + 77y - 133 = 0$
Now calculating the distance of the line $21x - 27y + 153 = 0$ from the origin
$d = \left| {\dfrac{{21 \cdot 0 - 27 \cdot 0 + 153}}{{\sqrt {{{21}^2} + {{27}^2}} }}} \right|$
$ = \left| {\dfrac{{153}}{{\sqrt {1170} }}} \right|$
Now calculating the distance of the line $99x + 77y - 133 = 0$ from the origin
$d = \left| {\dfrac{{99 \cdot 0 + 77 \cdot 0 - 133}}{{\sqrt {{{99}^2} + {{77}^2}} }}} \right|$
$ = \left| {\dfrac{{133}}{{\sqrt {15730} }}} \right|$
Since $\left| {\dfrac{{153}}{{\sqrt {1170} }}} \right| > \left| {\dfrac{{133}}{{\sqrt {15730} }}} \right|$. So, the line nearest to origin is $99x + 77y - 133 = 0$

Option ‘B’ is correct

Note: To find the locus of a point, you need to assume the coordinate of the point. Find the distance of the point from the given lines. Remember you need to consider the negative and positive of the distances. Since the lines are not parallel to each other. So that we get two angle bisectors. The distance of the lines from the angle bisector is always the same. That’s why you get two equations of the locus of the point.