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Question: One line passes through the point \[(1,9)\] and \((2,6)\) another line passes through \((3,3)\) and ...

One line passes through the point (1,9)(1,9) and (2,6)(2,6) another line passes through (3,3)(3,3) and (−1,5)( - 1,5). Then the acute angle between the two lines:
A. 300{30^0}
B. 450{45^0}
C. 600{60^0}
D. 1350{135^0}

Explanation

Solution

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(x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2}) is given by m=y2−y1x2−x1.m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}.
  • The angle θ\theta between two lines whose slopes are m1{m_1} and m2{m_2} is given by the formula:
    θ=tan⁡−1(m2−m11−m1m2)\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)(1,9) and (2,6)(2,6) be denoted by m1{m_1}.
Here, (x1,y1)=(1,9)({x_1},{y_1}) = (1,9) and (x2,y2)=(2,6)({x_2},{y_2}) = (2,6).
Substitute these values in slope formula m1=y2−y1x2−x1{m_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}
⇒m1=6−92−1\Rightarrow {m_1} = \dfrac{{6 - 9}}{{2 - 1}}
⇒m1=−31 ⇒m1=−3  \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)(3,3) and (−1,5)( - 1,5) be denoted by m2{m_2}.
Here, (x1,y1)=(3,3)({x_1},{y_1}) = (3,3) and (x2,y2)=(−1,5)({x_2},{y_2}) = ( - 1,5).
Substitute these values in slope formula m2=y2−y1x2−x1.{m_2} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}.
⇒m2=5−3−1−3\Rightarrow {m_2} = \dfrac{{5 - 3}}{{ - 1 - 3}}
⇒m2=2−4\Rightarrow {m_2} = \dfrac{2}{{ - 4}}
⇒m2=−12\Rightarrow {m_2} = - \dfrac{1}{2}
Find the angle between two lines.
Find the angle between two lines using the formula: θ=tan⁡−1(m2−m11−m1m2)\theta = {\tan ^{ - 1}}(\dfrac{{{m_2} - {m_1}}}{{1 - {m_1}{m_2}}}), put m1=−3,m2=−12{m_1} = - 3,{m_2} = - \dfrac{1}{2}
⇒θ=tan⁡−1[(−12−(−3))(1−(−12)(3))]\Rightarrow \theta = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{{ - 1}}{2} - ( - 3)} \right)}}{{\left( {1 - ( - \dfrac{1}{2})(3)} \right)}}} \right]
⇒θ=tan⁡−1[(−12+3)(1+32)]\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.
⇒θ=tan⁡−1[(−1+62)(2+32)] ⇒θ=tan⁡−1[(52)(52)]  \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] \\\
⇒θ=tan⁡−1(1)\Rightarrow \theta = {\tan ^{ - 1}}(1)
We know tan⁡(45∘)=1\tan ({45^ \circ }) = 1
⇒θ=tan⁡−1(tan⁡45∘)\Rightarrow \theta = {\tan ^{ - 1}}(\tan {45^ \circ })
Since, f−1(f(x))=x{f^{ - 1}}(f(x)) = x
⇒θ=450\Rightarrow \theta = {45^0}
Since, the angle is less than 90∘{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.