Question

the slope of the line joining the points (-5,4),(1,10) is tanθ.then express value of θ in values​

Answer 1

Equation of line joining (6,−2) and (−8,7)

y−(−2)=

−8−6

7−(−2)

(x−6)

y+2=−

14

9

(x−6)

14y+28=−9x+54

14y+9x=26.....(i)

Slope of line =m

1

=−

b

a

=−

14

9

2y−3x+7=0

Slope of line =m

2

=−

b

a

=−

2

−3

=

2

3

3x=2y+7

Substituting in (i)

14y+3(2y+7)=26

14y+6y+21=26

20y=5

⇒y=

4

1

3x=2(

4

1

)+7

3x=

2

15

⇒x=

2

5

So the point of intersection is (

2

5

,

4

1

)

Angle between two lines i.e. tanθ=

∣

∣

∣

∣

∣

1+m

1

m

2

m

1

−m

2

∣

∣

∣

∣

∣

tanθ=

∣

∣

∣

∣

∣

∣

∣

∣

∣

1+(−

14

9

)(

2

3

)

−

14

9

−

2

3

∣

∣

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

∣

28

28−27

28

−18−42

∣

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

∣

28

1

28

−60

∣

∣

∣

∣

∣

∣

∣

∣

tanθ=60

⇒θ=tan

−1

60

Answer 2

$Equation of line joining (6,−2) and (−8,7)</p><p>y−(−2)=−8−67−(−2)(x−6)y+2=−149(x−6)14y+28=−9x+5414yx=26.....(i)</p><p>Slope of line =m1=−ba=−149</p><p>2y−3x+7=0</p><p>Slope of line =m2=−ba=−2−3=23</p><p>3x=2y+7</p><p>Substituting in (i)</p><p>14y+3(2y+7)=2614y+6y+21=2620y=5⇒y=413x=2(41)+73x=215⇒x=25</p><p>So the point of intersection is (25,41Angle between two linesi.e. tanθ=∣∣∣∣∣1+m1m2m1−m2∣∣∣∣∣tanθ=∣∣∣∣∣∣∣∣∣1+(−149)(23)−149−23∣∣∣∣∣∣∣∣∣=∣∣∣∣∣∣∣∣2828−2728−18−42∣∣∣∣∣∣∣∣=∣∣∣∣∣∣∣∣28128−60∣∣∣∣∣∣∣∣tanθ=60⇒0$