Question

8. Find the equation of the straight line passing through the points
(i) (-2, 5) and (3, 6) (ii) (0, -6) and (-8, 2)

Answer 1

Hi ,

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We know that ,

Equation of a line passing through the points

( x1 , y1 ) and ( x2 , y2 ) is

( y - y1 )( x2 - x1 ) = ( x - x1 )( y2 - y1 )

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Now ,

i ) Let A( x1 , y1 ) = ( -2 , 5 ) ,

B( x2 , y2 ) = ( 3 , 6 )

Equation of a line AB is ,

( y - 5 )[ 3 - ( -2 ) ] = [ x - ( -2 )]( 6 - 5 )

( y - 5 )5 = ( x + 2 )1

5y - 25 = x + 2

0 = x - 5y +2 + 25

x - 5y + 27 = 0

ii ) Let P( x1 , y1 ) = ( 0 , 6 ) ,

Q( x2 , y2 ) = ( -8 , 2 )

Equation of the line PQ is ,

[ y - ( -6 ) ] ( -8 - 0 ) = ( x - 0 ) [ 2 - ( -6 ) ]

( y + 6 )( -8 ) = 8x

y + 6 = 8x/(-8 )

y + 6 = -x

x + y + 6 = 0

I hope this helps you.

: )