Math, asked by nizamuddinmd531, 11 months ago

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

Answers

Answered by MaheswariS
58

\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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Answered by isyllus
19

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.

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