Question

A variable line L is drawn through $O(0,0)$ to meet the lines ${L_1}:y - x - 10 = 0$ and ${L_2}:y - x - 20 = 0$ at the points A and B respectively. A point P is taken on L such that $\dfrac{2}{{OP}} = \dfrac{1}{{OA}} + \dfrac{1}{{OB}}$ and P, A, B lie on the same side of origin O. The locus of P is-
A) $3x + 3y - 40$
B) $3x + 3y + 40$
C) $3x - 3y - 40$
D) $3y - 3x = 40$

Answer 1

Hint- We will solve both the equations of the lines separately to find the coordinates of the intersection point A and B respectively. We will then find out the coordinates of point P in line L and then apply the distance formula in the equation given by the question i.e. $\dfrac{2}{{OP}} = \dfrac{1}{{OA}} + \dfrac{1}{{OB}}$.

Complete step-by-step answer:


The line L passes through origin at $O(0,0)$ . Thus, the equation of the line L will be-
$ \to L:y = mx$
Let the above equation be equation 1, we have-
 $ \to L:y = mx$ equation 1
The equation of the other two lines which are intersected by the line L at points A and B respectively is given to us by the question-
$
   \to {L_1}:y - x = 10 \\
    \\
   \to {L_2}:y - x = 20 \\
$
Now, we will find out the coordinates of points A and B on lines ${L_1}$ and ${L_2}$ respectively-
Point A: let the coordinates be $(x,y)$
Now, in the equation of line ${L_1}$, put the value of $y$ from equation 1 i.e. $y = mx$ and find the value of x, we get-
  $ \to x = \dfrac{{10}}{{m - 1}}$
Put this value of into the equation of line $L$ and find out the value of $y$ , we get-
$ \to y = \dfrac{{10m}}{{m - 1}}$
Thus, the coordinates of point A are:
$A = \left( {\dfrac{{10}}{{m - 1}},\dfrac{{10m}}{{m - 1}}} \right)$
Similarly, the coordinates of point B will be-
$B = \left( {\dfrac{{20}}{{m - 1}},\dfrac{{20m}}{{m - 1}}} \right)$
Let the coordinates of point P be $P(x,y)$.
Since point P lies on the line L, the coordinates of point P will be-
$P(x,mx)$ (Putting the value of y from equation 1)
Now, the equation given in the question is-
   $\dfrac{2}{{OP}} = \dfrac{1}{{OA}} + \dfrac{1}{{OB}}$ equation 2
Finding out the distance through distance formula $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ between OP, OA and OB keeping in mind that the coordinates of O are $(0,0)$ meaning the value of ${x_1}$ and ${y_1}$ is $(0,0)$-
For OP, ${x_2} = x,{y_2} = mx$
$OP = \sqrt {{x^2} + {m^2}{x^2}} $
For OA, ${x_2} = \dfrac{{10}}{{m - 1}},{y_2} = \dfrac{{10m}}{{m - 1}}$
$
  OA = \sqrt {{{\left( {\dfrac{{10}}{{m - 1}} - 0} \right)}^2} + {{\left( {\dfrac{{10m}}{{m - 1}} - 0} \right)}^2}} \\
    \\
   \Rightarrow \sqrt {{{\left( {\dfrac{{10}}{{m - 1}}} \right)}^2} + {{\left( {\dfrac{{10m}}{{m - 1}}} \right)}^2}} \\
$
For OB, ${x_2} = \dfrac{{20}}{{m - 1}},{y_2} = \dfrac{{20m}}{{m - 1}}$
$
  OB = \sqrt {{{\left( {\dfrac{{20}}{{m - 1}} - 0} \right)}^2} + {{\left( {\dfrac{{20m}}{{m - 1}} - 0} \right)}^2}} \\
    \\
   \Rightarrow \sqrt {{{\left( {\dfrac{{20}}{{m - 1}}} \right)}^2} + {{\left( {\dfrac{{20m}}{{m - 1}}} \right)}^2}} \\
$
Putting these values in equation 2, we get-
$
   \to \dfrac{2}{{\sqrt {{x^2} + {m^2}{x^2}} }} = \dfrac{1}{{\sqrt {{{\left( {\dfrac{{10}}{{m - 1}}} \right)}^2} + {{\left( {\dfrac{{10m}}{{m - 1}}} \right)}^2}} }} + \dfrac{1}{{\sqrt {{{\left( {\dfrac{{20}}{{m - 1}}} \right)}^2} + {{\left( {\dfrac{{20m}}{{m - 1}}} \right)}^2}} }} \\
    \\
   \Rightarrow \dfrac{2}{{x\sqrt {{m^2} + 1} }} = \dfrac{1}{{\dfrac{{10}}{{m - 1}}\sqrt {{m^2} + 1} }} + \dfrac{1}{{\dfrac{{20}}{{m - 1}}\sqrt {{m^2} + 1} }} \\
    \\
   \Rightarrow \dfrac{2}{x} = \dfrac{{m - 1}}{{10}} + \dfrac{{m - 1}}{{20}} \\
    \\
   \Rightarrow \dfrac{2}{x} = \dfrac{{2m - 2 + m - 1}}{{20}} = \dfrac{{3m - 3}}{{20}} \\
$
$ \Rightarrow $$40 = x\left( {3m - 3} \right)$ $ \to $ equation 3
Finding out the value of $m$ from equation 1, we get-
$m = \dfrac{y}{x}$
Putting this value of $m$ into equation 3-
$
   \to 40 = x\left( {\dfrac{{3y}}{x} - 3} \right) \\
    \\
   \Rightarrow \dfrac{{x\left( {3y - 3x} \right)}}{x} = 40 \\
    \\
   \Rightarrow 3y - 3x = 40 \\
$
Hence, the locus of point P is $3y - 3x = 40$.
Thus, option number D is correct.

Note: Finding out the distance between the points in the given equation in the question is necessary in order to put the values in the equation. Use the formula of distance $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $. Be clear about the coordinates of all the points.