Question

9. Find the equation of the straight line perpendicular to the straight line x - 2y +3 = 0 and passing through the point (1, -2).

Answer 1

Hi ,

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1 )Slope of a line ax + by + c = 0 is

m = - a/b

2 )slope of a line perpendicular to ax+by+c=0

is

m2 = b/a

3 ) slope - point form :

y - y1 = m( x - x 1 )

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Compare x - 2y + 3 = 0 with ax+by+c = 0

a = 1 , b = 2 , c = 3

slope of a line perpendicular to given line

is m = b/a = -2

This line passing through the point

( x1 , y1 ) = ( 1 , - 2 )


Equation of the line

y - y1 = m( x - x1 )

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

y + 2 = -2x + 2

y + 2 + 2x - 2 = 0

2x + y = 0

Therefore ,

Equation of a line perpendicular to

x - 2y +3 = 0 and passing through ( 1 , -2 )

is 2x + y = 0

I hope this helps you.

: )