Question: Find the separate equations of straight lines represented by the following equation: x² - 2xy cosecA + y² = 0.


Solution:
Given

Single equation of pair of straight lines is
x² - 2xy cosecA + y² = 0

[ Dividing by y² ]

or, $\dfrac{x²}{y²} - \dfrac{2xy cosecA}{y²} + \dfrac{y²}{y²} = 0$

or, $\left ( \dfrac{x}{y} \right )^2  - 2 cosecA \dfrac{x}{y} + 1 = 0$ - (i)

Comparing equation (i) with ax² + bx + c = 0
We get,
a = 1, b = -2 cosecA, c = 1, x = x/y


Using formula of quadratic equation,

$or, \dfrac{x}{y} = \dfrac{-b \pm \sqrt{b² - 4ac}}{2a} $

$= \dfrac{- (-2 cosecA) \pm \sqrt{(-2cosecA)² - 4×1×1}}{2}$

$= \dfrac{2 cosecA \pm \sqrt{4cosec²A - 4}}{2}$

$= \dfrac{2 cosecA \pm \sqrt{4(cosec²A - 1)}}{2}$

$= \dfrac{2 cosec\alpha \pm 2\sqrt{cosec²A - 1}}{2}$

[ cosec² A - 1 = cot² A ]

$= \dfrac{2 cosecA \pm 2\sqrt{cot² A}}{2}$

$or, \dfrac{x}{y} = \dfrac{2 cosecA \pm 2cotA}{2}$


Taking positive sign,

$or, \dfrac{x}{y} = \dfrac{2 cosecA + 2cotA}{2}$

$or, \dfrac{x}{y} = \dfrac{2 (cosecA+ cotA)}{2}$

$or, x = y(cosecA + cotA)$

$or, x = y cosecA +y cotA$

$or, x - y cosec A - y cotA = 0$

Taking negative sign,

$or, \dfrac{x}{y} = \dfrac{2 cosecA- 2cotA}{2}$

$or, \dfrac{x}{y} = \dfrac{2 (cosecA - cotA)}{2}$

$or, x = y(cosecA - cotA)$

$or, x = y cosecA - y cotA$

$or, x - y cosecA+ y cotA= 0$


Hence, the required separate equations of straight lines represented by the above equation is ($x - y cosec A - y cotA = 0$) and ($x - y cosecA+ y cotA= 0$).


Related Notes and Solutions:

Here is the website link to all the important formulae of Coordinate Geometry of Class 10.

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