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Question: The lines \(3x+y=0\) may be perpendicular to one of the lines given by \(a{{x}^{2}}+2hxy+b{{y}^{2}}=...

The lines 3x+y=03x+y=0 may be perpendicular to one of the lines given by ax2+2hxy+by2=0a{{x}^{2}}+2hxy+b{{y}^{2}}=0

Explanation

Solution

Here we have to find the condition of the required situation given. For that, we will first write the auxiliary equation. We will find the slope of the given line and then we will get the slope of the line perpendicular to it. We will put the value of slope in the auxiliary equation and will get the required conditions.

Complete step by step solution:
Let the auxiliary equation bm2+2hm+a=0\Rightarrow b{{m}^{2}}+2hm+a=0
Now, we will find the slope of the given line.
The equation of the line is 3x+y=03x+y=0
Therefore, the slope of given line is -3
Therefore, the slope of one of the lines represented by the joint equation will be equal to the negative of the inverse of the slope of the given line.
Slope=13=-\dfrac{1}{-3}
Therefore, the slope of one of the lines represented by the joint equation is 13\dfrac{1}{3}
13\dfrac{1}{3} must be the root of the auxiliary equation. So it will satisfy the equation.
We will put the value of slope in the auxiliary equation.
b(13)2+2h13+a=0b{{\left( \dfrac{1}{3} \right)}^{2}}+2h\dfrac{1}{3}+a=0
On further simplification, we get
b2+6h+9a=0{{b}^{2}}+6h+9a=0

This is the required condition.

Note:
We have calculated the slope of a given line in this question. We need to know its meaning.
A slope of a line represents the steepness of a line and it also represents the direction of the line.
If the value of slope is positive, then the line goes up from left to right and if the value of slope is negative then the line goes down from right to left. If the value of slope is zero then the line is horizontal and for vertical lines, the slope is undefined.