If the lines $y = 3 x + 7$ and $2 y + p x = 3$ are perpendicular to each other, find the value of $p$ .

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Solution

Step 1: Calculation of slope of line $y = 3 x + 7$ .
Given line is $y = 3 x + 7 - (1)$
The slope-intercept form of the equation of a line is given by $y = m x + c$ , where $m$ is the slope and $c$ is the $Y$ -intercept.
Comparing equation (1) with equation $y = m x + c$ , we get the slope of the line $y = 3 x + 7$ as $m_{1} = 3$ .
Step 2: Calculation of slope of line $2 y + p x = 3$ .
Simplify the equation $2 y + p x = 3$ to convert it into slope-intercept form.
$2 y + p x = 3 \Rightarrow 2 y = - p x + 3 (s u b t r a c t i n g p x f r o m b o t h s i d e s) ∴ y = - \frac{p}{2} x + \frac{3}{2} (d i v i d i n g b o t h s i d e s b y 2) - (2)$
Comparing equation (2) with equation $y = m x + c$ , we get the slope of the line $2 y + p x = 3$ as $m_{2} = - \frac{p}{2}$ .
Step3: Calculation of the value of $p$ .
Since, the lines $y = 3 x + 7$ and $2 y + p x = 3$ are perpendicular, the product of their slopes will be equal to $- 1$ .
$\Rightarrow m_{1} \cdot m_{2} = - 1 \Rightarrow 3 (- \frac{p}{2}) = - 1 \Rightarrow - \frac{3 p}{2} = - 1 \Rightarrow 3 p = 2 (m u l t i p l y i n g b o t h s i d e s b y - 2) ∴ p = \frac{2}{3} (d i v i d i n g b o t h s i d e s b y 3)$
Hence, the value of $p$ is $\frac{2}{3}$ .

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