Question

Locus of the points which are at equal distance from $3x+4y-11=0$ and $12x+5y+2=0$ and which is near the origin is

• A. $21x–77y+153=0$
• B. $99x+77y–133=0$
• C. $7x–11y=19$
• D. None of these

Answer 1

The correct option is B

$99x+77y–133=0$

Explanation for the correct option:

Step 1: Finding $d_1$ and $d_2$

Given the equation are

$3x+4y-11=0$ and

$12x+5y+2=0$

Finding the distance between the two line of equation using the formula,

$\Rightarrow d_1=\left|\frac{-ah-bk+c}{\sqrt{a^2+b^2}}\right|$

The given equations are of the form $ax+by+c=0$.

Comparing this equation with the given equation $3x+4y-11=0$

We get $a=3,b=4\&c=-11$

$\Rightarrow d_1=\left|\frac{-3h-4k-11}{\sqrt{9+16}}\right|$

Similarly, from equation $12x+5y+2=0$

We get $a=12,b=5\&c=2$

$\Rightarrow d_2=\left|\frac{12h+5k+2}{\sqrt{144+25}}\right|$

Step 2: Equating $d_1$ and $d_2$

Given, that the locus of the point is equal to the distance of the given equation.

Hence, equating $d_1\&d_2$

$$\begin{gathered} \Rightarrow \left|\frac{-3h-4k-11}{\sqrt{9+16}}\right|=\left|\frac{12h+5k+2}{\sqrt{144+25}}\right| \\ \Rightarrow \left|\frac{-3h-4k-11}{\sqrt{25}}\right|=\left|\frac{12h+5k+2}{\sqrt{169}}\right| \\ \Rightarrow \left|\frac{-3h-4k-11}{5}\right|=\left|\frac{12h+5k+2}{13}\right| \\ \Rightarrow 13\left(-3h-4k-11\right)=5(12h+5k+2) \\ \Rightarrow -39h-52k+143=60h+25k+10\mathbf{}\left(\mathrm{taking}+\mathrm{ve}\mathrm{sign}\mathrm{as}\mathrm{origin}\mathrm{is}\mathrm{contained}\mathrm{by}\mathrm{it}\right) \\ \Rightarrow 99h+77k-133=0 \end{gathered}$$

Therefore the locus of the points is $99x+77y–133=0$

Hence, option (B) is the correct answer.