IDNLearn.com: Where questions are met with accurate and insightful answers. Get the information you need from our experts, who provide reliable and detailed answers to all your questions.
Sagot :
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.
We are happy to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.