Question

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

Answer 1

Answer:

slope of a line ax+by+c=0 is

$m=\frac{-a}{b}$

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

$m_2=\frac{b}{a}$

slope point form

$y-y_1=m(x-x_1)$

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

$m=\frac{b}{a}=-2$

this line passing through the point

$(x_1,y_1)=(1,-2)$

equation of the line

$y-y_1=m(x-x_1)$

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

y+2=-2x+2

y+2+2x-2=0

2x+y=0

y=-2x

thus the required equation of line is y=-2x

Answer 2

Answer:

Answer:

slope of a line ax+by+c=0 is

m=\frac{-a}{b}m=

b

−a

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

m_2=\frac{b}{a}m

2

=

a

b

slope point form

y-y_1=m(x-x_1)y−y

1

=m(x−x

1

)

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=\frac{b}{a}=-2m=

a

b

=−2

this line passing through the point

(x_1,y_1)=(1,-2)(x

1

,y

1

)=(1,−2)

equation of the line

y-y_1=m(x-x_1)y−y

1

=m(x−x

1

)

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

y+2=-2x+2

y+2+2x-2=0

2x+y=0

y=-2x

thus the required equation of line is y=-2x