Question

If the slope of one of the lines represented by the equation \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] be \[\lambda \] times that of the other, then
A.\[4\lambda h=ab(1+\lambda )\]
B.\[\lambda h+ab{{(1+\lambda )}^{2}}\]
C.\[4\lambda {{h}^{2}}=ab{{(1+\lambda )}^{2}}\]
D.None of these

Answer 1

Hint: The slope of a line is the steepness and direction of a non-vertical line. When a line rises from left to right, the slope is positive. When a line falls from left to right, the slope is negative. If \[m\] represents the slope of a line and coordinates \[(x_1, y_1)\] and \[ x_1\] = \[ x_2\] respectively, then the slope of the line is given by the following formula.
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
 lf \[(x_1\] = \[X_2)\], then the line is vertical and the slope is undefined. Every line has an equation that can be written in the standard form \[Ax\text{ }+\text{ }By\text{ }=\text{ }C\] where\[A\], \[B\], and \[C\] are three integers, and \[A\] and B are not both zero. A must be positive.

Complete step-by-step answer:
Given that,
\[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\]
Let us assume that \[m\] is the slope.
Then according to the question other slope will be given as \[\lambda m\]
As we know that the sum of the slope is given as
\[{{m}_{1}}~+\text{ }{{m}_{2}}~=\dfrac{-2h}{b}\]
Using the above equation we get
\[m+\text{ }\lambda m~=\dfrac{-2h}{b}\]
Taking \[m\] as common factor from LHS we get
\[m(1+\text{ }\lambda )=\dfrac{-2h}{b}\]
Transposing we get
\[m=\dfrac{-2h}{b(1+\text{ }\lambda )}\]
As we know the product of the slope is given as
\[~{{m}_{1}}{{m}_{2}}~=\dfrac{a}{b}\]
Further equating we get
\[\lambda {{m}^{2}}~=\dfrac{a}{b}\]
Simplifying the equation we get
\[{{m}^{2}}~=\dfrac{a}{b\lambda }\]
Substituting the value of \[m\] in the above equation we get
\[\dfrac{4{{h}^{2}}}{{{b}^{2}}{{(1+\lambda )}^{2}}}=\dfrac{a}{b\lambda }\]
Rearranging the equation we get
\[4\lambda {{h}^{2}}=ab{{(1+\lambda )}^{2}}\]
Therefore, option \[C\] is the correct answer.
So, the correct answer is “Option C”.

Note: The order in which the points are taken really doesn't matter, as long as you subtract the x-values in the same order as you subtracted the y-values. A line is a curve in which every point on the line segment joining any two points on it lies on it.