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Find the angle between the two lines 3x+y+12=0 and x+2y−1=0. Find also the coordinates of their point of intersection and the equations of lines drawn perpendicular to them from the point (3,−2).

Easy

Solution

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3x+y+12=0......(i)

slope of line =m

1

=−

b

a

=−3

x+2y−1=0.......(ii)

slope of line =m

2

=−

b

a

=−

2

1

Angle between lines i.e. tanθ=

∣

∣

∣

∣

∣

1+m

1

m

2

m

1

−m

2

∣

∣

∣

∣

∣

tanθ=

∣

∣

∣

∣

∣

∣

∣

∣

∣

1+(−3)(−

2

1

)

−3−(−

2

1

)

∣

∣

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

∣

2

2+3

2

−6+1

∣

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

∣

2

5

2

−5

∣

∣

∣

∣

∣

∣

∣

∣

tanθ=1

⇒θ=45

∘

Now x+2y−1=0

x=1−2y

Substituting in (i)

3(1−2y)+y+12=0

3−6y+y+12=0

−5y=−15

⇒y=3

x=1−2(3)

⇒x=−5

So the point of intersection is (−5,3)

Let the slope of line perpendicular to (i) be m

m

1

m=−1

−3m=−1

⇒m=

3

1

It passes through (3,−2), so the equation of line is

y+2=

3

1

(x−3)

3y+6=x−3

x−3y=9

Let the slope of line perpendicualr to (ii) be m

′

−

2

1

m

′

=−1

⇒m

′

=2

It passes through (3,−2), so the equation of line is

y+2=2(x−3)

y+2=2x−6

2x−y=8

Answer:

3x+y+12=0......(i)

slope of line =m

1

=−

b

a

=−3

x+2y−1=0.......(ii)

slope of line =m

2

=−

b

a

=−

2

1

Angle between lines i.e. tanθ=

∣

∣

∣

∣

∣

1+m

1

m

2

m

1

−m

2

∣

∣

∣

∣

∣

tanθ=

∣

∣

∣

∣

∣

∣

∣

∣

∣

1+(−3)(−

2

1

)

−3−(−

2

1

)

∣

∣

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

∣

2

2+3

2

−6+1

∣

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

∣

2

5

2

−5

∣

∣

∣

∣

∣

∣

∣

∣

tanθ=1

⇒θ=45

∘

Now x+2y−1=0

x=1−2y

Substituting in (i)

3(1−2y)+y+12=0

3−6y+y+12=0

−5y=−15

⇒y=3

x=1−2(3)

⇒x=−5

So the point of intersection is (−5,3)

Let the slope of line perpendicular to (i) be m

m

1

m=−1

−3m=−1

⇒m=

3

1

It passes through (3,−2), so the equation of line is

y+2=

3

1

(x−3)

3y+6=x−3

x−3y=9

Let the slope of line perpendicualr to (ii) be m

′

−

2

1

m

′

=−1

⇒m

′

=2

It passes through (3,−2), so the equation of line is

y+2=2(x−3)

y+2=2x−6

2x−y=8

Step-by-step explanation:

3x+y+12=0......(i)

slope of line =m

1

=−

b

a

=−3

x+2y−1=0.......(ii)

slope of line =m

2

=−

b

a

=−

2

1

Angle between lines i.e. tanθ=

∣

∣

∣

∣

∣

1+m

1

m

2

m

1

−m

2

∣

∣

∣

∣

∣

tanθ=

∣

∣

∣

∣

∣

∣

∣

∣

∣

1+(−3)(−

2

1

)

−3−(−

2

1

)

∣

∣

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

∣

2

2+3

2

−6+1

∣

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

∣

2

5

2

−5

∣

∣

∣

∣

∣

∣

∣

∣

tanθ=1

⇒θ=45

∘

Now x+2y−1=0

x=1−2y

Substituting in (i)

3(1−2y)+y+12=0

3−6y+y+12=0

−5y=−15

⇒y=3

x=1−2(3)

⇒x=−5

So the point of intersection is (−5,3)

Let the slope of line perpendicular to (i) be m

m

1

m=−1

−3m=−1

⇒m=

3

1

It passes through (3,−2), so the equation of line is

y+2=

3

1

(x−3)

3y+6=x−3

x−3y=9

Let the slope of line perpendicualr to (ii) be m

′

−

2

1

m

′

=−1

⇒m

′

=2

It passes through (3,−2), so the equation of line is

y+2=2(x−3)

y+2=2x−6

2x−y=8