find the equation of the line perpendicular to the line x-2y+3=0 and passing through the point (1,-2).
Answers
Answer:
slope of a line ax+by+c=0 is
slope of a line perpendicular to ax+by+c=0 is
slope point form
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
this line passing through the point
equation of the line
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:
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