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Home » Find the angles between each of the following pairs of straight lines: (i) 3x + y + 12 = 0 and x + 2y – 1 = 0 (ii) 3x – y + 5 = 0 and x – 3y + 1 = 0
Find the angles between each of the following pairs of straight lines: (i) 3x + y + 12 = 0 and x + 2y – 1 = 0 (ii) 3x – y + 5 = 0 and x – 3y + 1 = 0
CBSE | Class 11 | Exercise 23.13 | Maths | RD Sharma | Straight Lines
Find the angles between each of the following pairs of straight lines: (i) 3x + y + 12 = 0 and x + 2y – 1 = 0 (ii) 3x – y + 5 = 0 and x – 3y + 1 = 0

(i) 

    \[3x\text{ }+\text{ }y\text{ }+\text{ }12\text{ }=\text{ }0\text{ }and\text{ }x\text{ }+\text{ }2y\text{ }\text{ }1\text{ }=\text{ }0\]

 According to ques,:

The equations of the lines are

    \[3x\text{ }+\text{ }y\text{ }+\text{ }12\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 1 \right)\]

    \[x\text{ }+\text{ }2y~-~1\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 2 \right)\]

Let m1 and m2 be the slopes of these lines.

    \[{{m}_{1}}~=\text{ }-3,\text{ }{{m}_{2}}~=\text{ }-1/2\]

Let θ be the angle between the lines.

Then, by using the formula

    \[tan\text{ }\theta \text{ }=\text{ }\left[ \left( {{m}_{1}}~\text{ }{{m}_{2}} \right)\text{ }/\text{ }\left( 1\text{ }+\text{ }{{m}_{1}}{{m}_{2}} \right) \right]\]

    \[=\text{ }\left[ \left( -3\text{ }+\text{ }1/2 \right)\text{ }/\text{ }\left( 1\text{ }+\text{ }3/2 \right) \right]\]

Or

    \[=\text{ }1\]

So,

    \[\theta \text{ }=\text{ }\pi /4\text{ }or\text{ }{{45}^{o}}\]

∴ The acute angle between the lines is 45°

    \[\left( \mathbf{ii} \right)~3x\text{ }\text{ }y\text{ }+\text{ }5\text{ }=\text{ }0\text{ }and\text{ }x\text{ }\text{ }3y\text{ }+\text{ }1\text{ }=\text{ }0\]

According to ques,:

The equations of the lines are

    \[3x~-~y\text{ }+\text{ }5\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 1 \right)\]

    \[x~-~3y\text{ }+\text{ }1\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 2 \right)\]

Let m1 and m2 be the slopes of these lines.

    \[{{m}_{1}}~=\text{ }3,\text{ }{{m}_{2}}~=\text{ }1/3\]

Let θ be the angle between the lines.

Then, by using the formula

    \[tan\text{ }\theta \text{ }=\text{ }\left[ \left( {{m}_{1}}~\text{ }{{m}_{2}} \right)\text{ }/\text{ }\left( 1\text{ }+\text{ }{{m}_{1}}{{m}_{2}} \right) \right]\]

    \[=\text{ }\left[ \left( 3\text{ }+\text{ }1/3 \right)\text{ }/\text{ }\left( 1\text{ }+\text{ }1 \right) \right]\]

Or

    \[=\text{ }4/3\]

So,

    \[\theta \text{ }=\text{ }ta{{n}^{-1}}~\left( 4/3 \right)\]

∴ The acute angle between the lines is tan-1 (4/3).

i

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