Answer :
given: 2x-y-3=0.
find equation for the line perpendicular to the given line that goes through the given point:
(2;-1)koord of direction vector (i`m not know how it is called at you, because i'm from russia)
=> (x-0)/2=(y-4)/-1 (canonical equation)
=>x+2y-8=0(general equation)
further:
{x+2y-8=0
{2x-y-3 =0 => y=13/5 x=14/5
(14/5; 13/5) - koord point on line
|dist|=sqrt( (14/5-0)^2 + (13/5-4)^2 ) = sqtr(7.72) = 2.78
Удачи!
.
find equation for the line perpendicular to the given line that goes through the given point:
(2;-1)koord of direction vector (i`m not know how it is called at you, because i'm from russia)
=> (x-0)/2=(y-4)/-1 (canonical equation)
=>x+2y-8=0(general equation)
further:
{x+2y-8=0
{2x-y-3 =0 => y=13/5 x=14/5
(14/5; 13/5) - koord point on line
|dist|=sqrt( (14/5-0)^2 + (13/5-4)^2 ) = sqtr(7.72) = 2.78
Удачи!
.
the line y = 2x -3 has a gradient of 2.
so the line perpendicular to it has a gradient of -1/2 and will have the general formula of y = -1/2 x + c.
to find c, use the coordinates (0,4)
4 = 0 + c
so c = 4
equation is y = -1/2 x + 4
If you need the distance of (0,4) from the line you will need to put both lines on a graph to find the intersection (and possibly use simultaneous equations for more accurate answer) and then use pythagoras to find lengths.
so the line perpendicular to it has a gradient of -1/2 and will have the general formula of y = -1/2 x + c.
to find c, use the coordinates (0,4)
4 = 0 + c
so c = 4
equation is y = -1/2 x + 4
If you need the distance of (0,4) from the line you will need to put both lines on a graph to find the intersection (and possibly use simultaneous equations for more accurate answer) and then use pythagoras to find lengths.