Question

4. Find the angle of inclination of the line passing through the points
(i) (1,2) and (2,3) (ii) (3,3) and (0,0) (iii) (a , b) and (-a , -b)

Answer 1

Let θ be the angle of inclination of the line.
Then, slope of the line, m = tan θ
Also we know, slope of two given points (x₁, y₁) and (x₂,y₂) is (y₂ -y₁)/(x₂ - x₁)
so, (y₂ - y₁)/(x₂ - x₁) = tanθ

(i) (1,2) and (2,3)
slope of line joining given points = (3 - 2)/(2 - 1) = 1
1 = tanθ = tan45°
θ = 45°

(ii) (3,3) and (0,0)
slope = (0 - 3)/(0 -3) = 1
1 = tanθ = tan45°
θ = 45°

(iii)(a,b) and (-a, -b)
slope = (-b - b)/(-a - a) = -2b/-2a = b/a
tanθ = b/a
θ = arctan(b/a)

Answer 2

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Slope of the line containing

the points ( x1 , y1 ) and (x2 ,y2 )

is m = ( y2 - y1 )/( x2 - x1 )

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Here ,

i ) A( 1 , 2 ) = ( x1 , y1 )

B( 2 , 3 ) = ( x2 , y2 ),

Slope of line AB = (3 - 2)/(2-1)

m = 1

ii ) P( 3 , 3 ) = ( x1 , y1 ),

Q( 0 , 0 ) = ( x2 , y2 ) ,

Slope of a line PQ

= ( 0 - 3 )/( 0 - 3 )

= ( -3 )/( -3 )

m = 1

iii ) C( a , b ) = ( x1 , y1 ),

D( -a , - b ) = ( x2 , y2 )

Slope of a line CD

= ( -b - b )/( -a - a )

m = ( -2b )/( -2a )

m = b/a

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