Question

2.
Find the value of p, if the straight lines 3x + 7y - 1 = 0 and
7x - py + 3 = 0 are mutually perpendicular.​

Answer 1

$\textbf{Given:}$

$3x+7y-1=0\;\text{and}\;7x-py+3=0$

$\text{are mutually perpendicular}$

$\textbf{To find:}$

$\text{The value of p}$

$\textbf{Solution:}$

$\textbf{Slope of the line ax+by+c=0 is}\;m=\dfrac{-a}{b}$

$\text{Slope of the line 3x+7y-1=0 is}\;m_1=\dfrac{-3}{7}$

$\text{Slope of the line 7x-py+3=0 is}\;m_2=\dfrac{-7}{-p}=\dfrac{7}{p}$

$\text{Since the lines are perpendicular, we have}$

$\bf\,m_1{\times}m_2=-1$

$\dfrac{-3}{7}{\times}\dfrac{7}{p}=-1$

$\dfrac{-3}{p}=-1$

$\dfrac{3}{p}=1$

$\implies\,p=3$

$\textbf{Answer:}$

$\textbf{The value of p is 3}$

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2.For what value of a the system of linear equations 2x + 3y = 7 and ( a -1 )x + (a + 1)y = 3a + 1 represent parallel lines?​

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Answer 2

Given:

Two straight lines:

$3x + 7y - 1 = 0$ and

$7x - py + 3 = 0$

These are perpendicular to each other.

To find:

Value of p = ?

Solution:

Let us first learn the concept of perpendicular lines.

When two lines with slopes $m_1$ and $m_2$ are perpendicular to each other then the following condition holds true:

$m_1 \times m_2 = -1$

Let us convert the given straight line equations to slope intercept forms.

Slope intercept form is given as:

$y=mx+c$

1st equation:

$3x + 7y - 1 = 0\\\Rightarrow 7y = -3x+1\\\Rightarrow y = -\dfrac{3}{7}x+\dfrac{1}{7}$

$\therefore m_1 = -\dfrac{3}{7}$

2nd equation:

$7x - py + 3 = 0\\\Rightarrow py = 7x+3\\\Rightarrow y = \dfrac{7}{p}+\dfrac{3}{p}$

$\therefore m_2 = -\dfrac{7}{p}$

The lines are given to be mutually perpendicular:

$\therefore m_1 \times m_2 = -1\\\Rightarrow -\dfrac{3}{7} \times \dfrac{7}{p} = -1\\\Rightarrow \dfrac{3}{p} = 1\\\Rightarrow p = 3$

So, the values of p is 3.