Question: Lines $(a-1)x+y=5$ and $(a+1)x -3y=15$ are perpendicular to each other. Find the value of a.


Solution:
Given,

Equation of line (i) is $(a-1)x +y = 5$

Slope of line (i), $(m_1) = - \dfrac{coefficient\; of \;x}{coefficient\;of\;y}$
$= - \dfrac{a-1}{1}$

Equation of line (ii) is $(a+1)x -3y)= 15$

Slope of line (ii), $(m_2) = - \dfrac{coefficient\; of \;x}{coefficient\;of\;y}$
$= - \dfrac{a+1}{-3}$
$= \dfrac{a+1}{3}$

We know,
When two lines are perpendicular, the product of their slope is equal to negative 1.

$or, m_1 × m_2 = -1$
$or, - \dfrac{a-1}{1} × \dfrac{a+1}{3} = -1$

$or, - \dfrac{(a-1)(a+1)}{3} = -1$

$or, a² - 1 = 3$

$or, a² = 3+1$

$or, a² = 4$

$or, a = \pm \sqrt{4}$

$\therefore a = \pm 2$

Hence, the value of a is either 2 or -2.

Related Notes and Solutions:

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

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