Find the slopes $m_{1}$ and $m_{2}$ of the straight lines $y = 3 x - 2$ and $x + 3 y + 5 = 0$
Calculate $m_{1} \cdot m_{2}$
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Solution

In the equation $y = 3 x - 2$ the slope will be 3 because it is in the form of $y = m x + c$ .

In the equation $x + 3 y + 5 = 0$ , let's take the y-term to one side and rest of the terms to other side.
$⟹ 3 y = - x - 5 ⟹ y = - \frac{1}{3} x - \frac{5}{3}$
So, the slope here is $\frac{- 1}{3}$ .

Now, we calculate the multiplication $m_{1} \cdot m_{2} = 3 \cdot (- \frac{1}{3})$

$m_{1} \cdot m_{2} = - 1$

Whenever the product of slopes of two lines are -1 then the lines are perpendicular to each other.

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Find the slopes m1 and m2 of the straight lines y=3x−2 and x+3y+5=0 Calculate m1⋅m2 What do you perceive?

Find the slopes $m_{1}$ and $m_{2}$ of the straight lines $y = 3 x - 2$ and $x + 3 y + 5 = 0$
Calculate $m_{1} \cdot m_{2}$
What do you perceive?