
One line passes through the point \[(1,9)\] and $(2,6)$ another line passes through $(3,3)$ and $( - 1,5)$. Then the acute angle between the two lines:
A. ${30^0}$
B. ${45^0}$
C. ${60^0}$
D. ${135^0}$
Answer
484.5k+ views
Hint: Here we use the points given to find slopes of two different lines and then using the formula for angle between the lines we can find the measure of the angle.
* The slope of line passing through two points$({x_1},{y_1})$ and $({x_2},{y_2})$ is given by $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}.$
* The angle $\theta $ between two lines whose slopes are ${m_1}$ and ${m_2}$ is given by the formula:
$\theta = {\tan ^{ - 1}}(\dfrac{{{m_2} - {m_1}}}{{1 - {m_1}{m_2}}})$.
Complete step-by-step answer:
Find the slope of the first line.
Let the slope of the line passing through\[(1,9)\] and $(2,6)$ be denoted by ${m_1}$.
Here, $({x_1},{y_1}) = (1,9)$ and $({x_2},{y_2}) = (2,6)$.
Substitute these values in slope formula ${m_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
$ \Rightarrow {m_1} = \dfrac{{6 - 9}}{{2 - 1}}$
$
\Rightarrow {m_1} = \dfrac{{ - 3}}{1} \\
\Rightarrow {m_1} = - 3 \\
$
Find the slope of the second line.
Let the slope of the line passing through \[(3,3)\] and $( - 1,5)$ be denoted by ${m_2}$.
Here, $({x_1},{y_1}) = (3,3)$ and $({x_2},{y_2}) = ( - 1,5)$.
Substitute these values in slope formula ${m_2} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}.$
$ \Rightarrow {m_2} = \dfrac{{5 - 3}}{{ - 1 - 3}}$
$ \Rightarrow {m_2} = \dfrac{2}{{ - 4}}$
$ \Rightarrow {m_2} = - \dfrac{1}{2}$
Find the angle between two lines.
Find the angle between two lines using the formula: $\theta = {\tan ^{ - 1}}(\dfrac{{{m_2} - {m_1}}}{{1 - {m_1}{m_2}}})$, put ${m_1} = - 3,{m_2} = - \dfrac{1}{2}$
$ \Rightarrow \theta = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{{ - 1}}{2} - ( - 3)} \right)}}{{\left( {1 - ( - \dfrac{1}{2})(3)} \right)}}} \right]$
$ \Rightarrow \theta = {\tan ^{ - 1}}\left[ {\dfrac{{\left( { - \dfrac{1}{2} + 3} \right)}}{{\left( {1 + \dfrac{3}{2}} \right)}}} \right]$
Solving both numerator and denominator by taking LCM separately.
$
\Rightarrow \theta = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{{ - 1 + 6}}{2}} \right)}}{{\left( {\dfrac{{2 + 3}}{2}} \right)}}} \right] \\
\Rightarrow \theta = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{5}{2}} \right)}}{{\left( {\dfrac{5}{2}} \right)}}} \right] \\
$
$ \Rightarrow \theta = {\tan ^{ - 1}}(1)$
We know \[\tan ({45^ \circ }) = 1\]
$ \Rightarrow \theta = {\tan ^{ - 1}}(\tan {45^ \circ })$
Since, \[{f^{ - 1}}(f(x)) = x\]
$ \Rightarrow \theta = {45^0}$
Since, the angle is less than \[{90^ \circ }\], therefore it is an acute angle.
So, the correct answer is “Option B”.
Note: Students many times make mistakes while finding the value of angle, they should always write the value inside the inverse function as the function and then cancel the inverse function with the function. Also, keep in mind slope of a line should be in simplest form, i.e. there should be no common factor between the numerator and denominator.
* The slope of line passing through two points$({x_1},{y_1})$ and $({x_2},{y_2})$ is given by $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}.$
* The angle $\theta $ between two lines whose slopes are ${m_1}$ and ${m_2}$ is given by the formula:
$\theta = {\tan ^{ - 1}}(\dfrac{{{m_2} - {m_1}}}{{1 - {m_1}{m_2}}})$.
Complete step-by-step answer:
Find the slope of the first line.
Let the slope of the line passing through\[(1,9)\] and $(2,6)$ be denoted by ${m_1}$.
Here, $({x_1},{y_1}) = (1,9)$ and $({x_2},{y_2}) = (2,6)$.
Substitute these values in slope formula ${m_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
$ \Rightarrow {m_1} = \dfrac{{6 - 9}}{{2 - 1}}$
$
\Rightarrow {m_1} = \dfrac{{ - 3}}{1} \\
\Rightarrow {m_1} = - 3 \\
$
Find the slope of the second line.
Let the slope of the line passing through \[(3,3)\] and $( - 1,5)$ be denoted by ${m_2}$.
Here, $({x_1},{y_1}) = (3,3)$ and $({x_2},{y_2}) = ( - 1,5)$.
Substitute these values in slope formula ${m_2} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}.$
$ \Rightarrow {m_2} = \dfrac{{5 - 3}}{{ - 1 - 3}}$
$ \Rightarrow {m_2} = \dfrac{2}{{ - 4}}$
$ \Rightarrow {m_2} = - \dfrac{1}{2}$
Find the angle between two lines.
Find the angle between two lines using the formula: $\theta = {\tan ^{ - 1}}(\dfrac{{{m_2} - {m_1}}}{{1 - {m_1}{m_2}}})$, put ${m_1} = - 3,{m_2} = - \dfrac{1}{2}$
$ \Rightarrow \theta = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{{ - 1}}{2} - ( - 3)} \right)}}{{\left( {1 - ( - \dfrac{1}{2})(3)} \right)}}} \right]$
$ \Rightarrow \theta = {\tan ^{ - 1}}\left[ {\dfrac{{\left( { - \dfrac{1}{2} + 3} \right)}}{{\left( {1 + \dfrac{3}{2}} \right)}}} \right]$
Solving both numerator and denominator by taking LCM separately.
$
\Rightarrow \theta = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{{ - 1 + 6}}{2}} \right)}}{{\left( {\dfrac{{2 + 3}}{2}} \right)}}} \right] \\
\Rightarrow \theta = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{5}{2}} \right)}}{{\left( {\dfrac{5}{2}} \right)}}} \right] \\
$
$ \Rightarrow \theta = {\tan ^{ - 1}}(1)$
We know \[\tan ({45^ \circ }) = 1\]
$ \Rightarrow \theta = {\tan ^{ - 1}}(\tan {45^ \circ })$
Since, \[{f^{ - 1}}(f(x)) = x\]
$ \Rightarrow \theta = {45^0}$
Since, the angle is less than \[{90^ \circ }\], therefore it is an acute angle.
So, the correct answer is “Option B”.
Note: Students many times make mistakes while finding the value of angle, they should always write the value inside the inverse function as the function and then cancel the inverse function with the function. Also, keep in mind slope of a line should be in simplest form, i.e. there should be no common factor between the numerator and denominator.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
