Question

From a point $A\left( {1,1} \right)$ straight lines AL and AM are drawn at right angles to the pair of straight lines $3{x^2} + 7xy - 2{y^2} = 0$. Find the equation of pair straight lines AL and AM.

Answer 1

In this particular question use the concept that first find out the perpendicular on the given pair of straight lines by using the concept that if $a{x^2} + 2hxy + b{y^2} = 0$ is the pair of straight lines then the equation of perpendicular is constructed by interchanging the coefficients of ${x^2}{\text{ and }}{y^2}$ and multiply by negative in the coefficient of XY, then shift the axis of this perpendicular equation so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given data:
From a point $A\left( {1,1} \right)$ straight lines AL and AM are drawn at right angles to the pair of straight lines $3{x^2} + 7xy - 2{y^2} = 0$.
We have to find out the equation of AL and AM.
Now as we know that if $a{x^2} + 2hxy + b{y^2} = 0$ is the pair of straight lines then the equation of perpendicular is constructed by interchanging the coefficients of ${x^2}{\text{ and }}{y^2}$ and multiply by negative in the coefficient of xy.
So the line which is perpendicular to $a{x^2} + 2hxy + b{y^2} = 0$ is $b{x^2} - 2hxy + a{y^2} = 0$.
So, the line perpendicular to line $3{x^2} + 7xy - 2{y^2} = 0$ is, here a = 3, 2h = 7, b = -2
$\Rightarrow - 2{x^2} - 7xy + 3{y^2} = 0$............. (1)
So equation (1) is perpendicular to $3{x^2} + 7xy - 2{y^2} = 0$.
Now we have to find out the equation passing from point A (1, 1) perpendicular to $3{x^2} + 7xy - 2{y^2} = 0$.
So if we shift the axis of equation (1) by (1, 1) we will get the equation of line perpendicular to $3{x^2} + 7xy - 2{y^2} = 0$ and passing from point (1, 1).
So, replace $x \to x - 1,y \to y - 1$ in equation (1) we have,
$\Rightarrow - 2{\left( {x - 1} \right)^2} - 7\left( {x - 1} \right)\left( {y - 1} \right) + 3{\left( {y - 1} \right)^2} = 0$
Now simplify it we have,
$\Rightarrow - 2\left( {{x^2} + 1 - 2x} \right) - 7\left( {xy - x - y + 1} \right) + 3\left( {{y^2} + 1 - 2y} \right) = 0$
$\Rightarrow - 2{x^2} + 3{y^2} - 7xy + 11x + y - 6 = 0$
Multiply by -1 throughout we have,
$\Rightarrow 2{x^2} - 3{y^2} + 7xy - 11x - y + 6 = 0$
So this is the required equation passing from point A (1, 1) and represents a pair of straight lines AM and AL.
So this is the required answer.

Note: Whenever we face such types of equations the key concept we have to remember is how to shift the axis of a straight line i.e. if we wanted to shift the axis by (h, k) so replace $x \to x - h$ and $y \to y - k$ or point is (1, 1) so replace $x \to x - 1,y \to y - 1$ and solve as above we will get the required answer.