Question

If the graph of 2x+3y − 6=0 is perpendicular to the graph of ax − 3y=5. What is the value of a?

Answer 1

Answer:

$a = \frac{9}{2}$

Step-by-step explanation:

To find the value of a,

Find the Slope of both the equation

For lines to be perpendicular to each other

$\boxed{m_1m_2 = - 1}$

For line 1:2x+3y − 6=0

represent the line in y=mx +c form

$3y = - 2x + 6 \\ \\ y = \frac{ - 2}{3}x + \frac{6}{3} \\ \\ y = \frac{ - 2x}{3} + 2 \\ \\ m_1 = \frac{ - 2}{3}$

For line 2:

$ax - 3y = 5 \\ \\ - 3y = - ax + 5 \\ \\ 3y = ax - 5 \\ \\ y = \frac{ax}{3} - \frac{5}{3} \\ \\ m_2 = \frac{a}{3}$

Apply the condition of perpendicularity

$\frac{ - 2}{3} \times \frac{a}{3} = - 1 \\ \\ \frac{2a}{9} = 1 \\ \\ a = \frac{9}{2}$

Hope it helps you.

Answer 2

concept : as you angle between two lines is given as $tan\theta=\left|\frac{m_1-m_2}{1+m_1.m_2}\right|$

where, $m_1$ and $m_2$ are slopes of given lines.

if lines are perpendicular to each other.

then, θ = 90°

so, tan90° = $tan\theta=\left|\frac{m_1-m_2}{1+m_1.m_2}\right|$

⇒1/0 = $tan\theta=\left|\frac{m_1-m_2}{1+m_1.m_2}\right|$

⇒1 + $m_1.m_2$ = 0

⇒$\bf{m_1.m_2}$ = - 1, this is required condition when two lines are perpendicular to each other.

given lines : 2x + 3y - 6 = 0 , ax - 3y = 5

slope of first line, $m_1$ = -2/3

slope of 2nd line, $m_2$ = a/3

now, applying $m_1.m_2=-1$

⇒-2/3 × a/3 = -1

⇒-2/9a = -1

⇒ a = 9/2

hence, value of a = 9/2