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Maths-

General

Easy

Question

The angle between 2x + 3y – 7 = 0 and x + y – 8 = 0 is

$\tan^{- 1} (\frac{3}{4})$
$\tan^{- 1} (\frac{3}{5})$
tan^–1 (5)
$\tan^{- 1} (\frac{1}{5})$

hint Hint:

First, we will find the slopes of the lines. The slope of a line $a x plus b y plus c equals 0$ is given by $m equals fraction numerator negative c o e f f i c i e n t space o f space y space over denominator c o e f f i c i e n t space o f space x end fraction$ . After that we will use the formula of the angle between the two lines $tan open parentheses theta close parentheses equals open vertical bar fraction numerator m subscript 1 minus m subscript 2 over denominator 1 plus m subscript 1 m subscript 2 end fraction close vertical bar$ where $m subscript 1$ and $m subscript 2$ are the slope of the lines and $theta$ is the angle between the lines.

The correct answer is: $tan to the power of negative 1 end exponent invisible function application open parentheses 1 fifth close parentheses$

The formula of the angle between the two lines is given by $tan open parentheses theta close parentheses equals open vertical bar fraction numerator m subscript 1 minus m subscript 2 over denominator 1 plus m subscript 1 m subscript 2 end fraction close vertical bar$ where $m subscript 1$ and $m subscript 2$ are the slope of the lines and $theta$ is the angle between the lines.
Step1: Finding the Slopes of the lines.
The slope of a line $a x plus b y plus c equals 0$ is given by $m equals fraction numerator negative c o e f f i c i e n t space o f space y space over denominator c o e f f i c i e n t space o f space x end fraction$ hence, $m equals fraction numerator negative b over denominator a end fraction$ .
Therefore slopes of the given lines are $m subscript 1 equals fraction numerator negative 3 over denominator 2 end fraction$ and $m subscript 2 equals fraction numerator negative 1 over denominator 1 end fraction$
Step2: Finding the angles between the lines.
$tan open parentheses theta close parentheses equals open vertical bar fraction numerator m subscript 1 minus m subscript 2 over denominator 1 plus m subscript 1 m subscript 2 end fraction close vertical bar$
=> $tan open parentheses theta close parentheses equals open vertical bar fraction numerator begin display style fraction numerator negative 3 over denominator 2 end fraction end style minus left parenthesis begin display style fraction numerator negative 1 over denominator 1 end fraction end style right parenthesis over denominator 1 plus left parenthesis begin display style fraction numerator negative 3 over denominator 2 end fraction end style right parenthesis cross times begin display style fraction numerator negative 1 over denominator 1 end fraction end style end fraction close vertical bar$
=> $tan open parentheses theta close parentheses equals open vertical bar fraction numerator begin display style fraction numerator negative 3 over denominator 2 end fraction plus 1 end style over denominator 1 plus left parenthesis begin display style fraction numerator plus 3 over denominator 2 end fraction end style right parenthesis cross times begin display style 1 end style end fraction close vertical bar$
=> $tan open parentheses theta close parentheses equals open vertical bar fraction numerator begin display style fraction numerator negative 3 plus 2 over denominator 2 end fraction end style over denominator begin display style fraction numerator 2 plus 3 over denominator 2 end fraction end style end fraction close vertical bar$
=> $tan open parentheses theta close parentheses equals open vertical bar fraction numerator begin display style negative 3 plus 2 end style over denominator begin display style 2 plus 3 end style end fraction close vertical bar$
=> $tan open parentheses theta close parentheses equals open vertical bar fraction numerator begin display style negative 1 end style over denominator begin display style plus 5 end style end fraction close vertical bar$
=> $tan open parentheses theta close parentheses equals open vertical bar 1 fifth close vertical bar equals greater than 1 fifth$
=> So, $theta equals tan to the power of negative 1 end exponent open parentheses 1 fifth close parentheses$

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