Find the point P on the line Y=3x???
y=3x slope = 3 when slope is negative reciprocal (-1/3) the line perpendicular y = -1/3(x-50) = -1/3 x + 50/3 3x = -1/3 x + 5/3 x = 5 y = 15 P(5, 15) d = √[(50-5)^2+(0-15)^2] = 15√10
Find point P on the line y=5x that is closest to the point (52,0). What is the least distance between P and (52,0)?
d=√)(x-52)^2+25x^2) You find the minimum
Let A = (-2, 4) and B = (7, 6). Find the point P on the line y=2 that makes the total distance AP + BP as small as possible.?
Will someone please explain this problem to me? Or at least the methodology? Do you have to take the perpendicular bisector s intersection with the line, or find the derivative of something? Please help, thanks. PS: I m not trying to cheat on HW, I m actually doing this for fun and just really want to figure it out. Thanks!
Find the point P where the line x = 1 + t, y = 2t, z = -3t intersects the plane x + y - z = 1.?
Here is how: plug 1+t in for x and 2t in for y and -3t in for z in the equation x + y - z = 1. This will give you a single equation for t. Solve it. Take that numeric value of t and plug it into (x,y,z) = (1+t, 2t, -3t) and you will have your point.
Optimization: Find the point P on the line y=2x that is closest to the point (20,0). What is the least distance between P and (20,0)?
One way to determine this: Calculus minimize y = sqrt ( (x-20)^2 + 4x^2 ) = ( x^2 - 40x + 400 + 4x^2) ^(1/2) y = (5x^2 - 40x + 400)^(1/2) dy/dx = (10x - 40)/ (5x^2 -40x + 400)^(1/2) so minimized or maximized when dy/dx = 0 Since the numerator 5x^2 -40x + 400 is always greater than 0 then dy/dx is only zero when the numerator is 0 10x -40 = 0 10x = 40 x = 40/10 x = 4 y = 2x y = 2*4 = 8 By this method , c (4,8) == Method two Method two use Geometry, The shortest distance between a line and point is given by a perpendicular line SO you want a line with slope -1/2 = -1/m = -1/ (2) = -1/2 Line y = (-1/2)x + b but it goes through point (20,0) so b = (1/2)x + y = (1/2)*20 + 0 = 10 y = (-1/2)x + 10 So the shortest distance is where the two perpendicular lines intersect. y = (-1/2)x + 10 and y = 2x intersect when 2x = (-1/2)x + 10 2x = (-1/2)x + 10 2x + 0.5x = 10 2.5x = 10 x = 4 y = 2x = 2*4 = 8 so by this method also, answer to C. (4, 8) c. (4,8) d. use b dist = sqrt((4 -20)^2 + 4*4^2 )) = sqrt (16^2 + 4*16) = sqrt(256 + 64) = sqrt( 320) =sqrt(64)*sqrt(5) dist = 8*sqrt(5) = approx. 17.8885 d. (17.8885)
Find the point P where the line x = 1 + t, y = 2t, z = -3t intersects the plane x + y - z = 4.?
If you want the number for t here's the answer: Place "1+t"instead of "x" "2t" instead of "y" And "-3t" instead of "z" So you should just solve a simple equation: (1+t)+(2t)-(-3t)=4 1+3t+3t=4 1+6t=4 then decrease 1 from both sides of equation We have: 6t=3 =>t=0.5
How can one find the equation of the locus of the point P, if the points A (1, 2), B (3, 4) are at an equal distance from P?
First of all, notice that every point on locus of P will be at equal distance from A(1,2) and B(3,4). So the locus will be perpendicular bisector of AB. Let us proove this now.We will follow the general procedure. Let the point P be (h,k). Applying distance formula on AP=PB , we get,(h-1)^2 + (k-2)^2 = (h-3)^2 + (k-4)^2On solving,1–2h+4–4k = 9–6h+16–8k4h+4k = 20h+k = 5.Now replacing h by x, and k by y, we get,x+y = 5.This is the required locus.Also the midpoint of A and B is (2,3) and slope of AB is 1.The locus passes through (2,3) and its slope is -1. Thus it is the perpendicular bisector of AB.