Question: Prove that the equation 4x² -15xy -4y² = 0 represents two straight lines through origin which are perpendicular to each other.

Solution:

Given,
Single equation of two lines is $4x² -15xy -4y² = 0$
$or, 4x² -15xy -4y² = 0$
$or, 4x² -(16-1)xy -4y² = 0$
$or, 4x² -16xy +xy -4y² = 0$
$or, 4x(x -4y) + y(x -4y) = 0$
$or, (4x +y)(x -4y) = 0$

So, the required two equations of straight lines are $4x +y = 0$ and $x -4y = 0$.
Since, the above two lines do not have any y-intercepts it is proved that they lie through origin.

Now,
For line 1 (4x + y = 0),
Slope ($m_1$) =$ \frac{- coefficient \; of \; x}{coefficient \; of \;y}$
$= \frac{-4}{1}$
$= -4$

And,
For line 2 (x -4y = 0),
Slope ($m_2$) = $\frac{-coefficent \; of \; x}{coefficient \; of \; y}$
$= \frac{- 1}{-4}$
$= \frac{1}{4}$

For two straight lines to be perpendicular, their slopes should be in the following relation,
$m_1 × m_2 = -1$
$or, -4 × \frac{1}{4} = -1$
$or, -1 = -1$ which is true.
So, the lines are perpendicular to each other.

Hence, it is proved that the two lines represented by the given equation of single line are passing through the origin and are perpendicular to each other.

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