Pick an arbitrary point (x,y). What is the distance from (x,y) and point P? Is there a constraint on the point (x,y)? If yes, include that constraint into the distance from (x,y) to point P.
Please show us your work in finding the distance from (x,y) to point P.
That's nice, but are you sure this isn't in a textbook section called "Applications of the 1st Derivative"?In year 12 not really calculus xD
yuh so Speed * Time = Distance
so square root of (4-10)^2+(5-2)^2
comes out to be square root of 45
You certainly can find the distance between (x,y) and (10,2). It is \(\displaystyle d=\sqrt{(x-10)^2-(y-2)^2)}\). Now the constraint is y=2x-3 so just update that distance formula and get \(\displaystyle d=\sqrt{(x-10)^2-(2x-3-2)^2)}\).well i couldnt figure out a way to do it with calculus but if u kniw that a perpenidcular line is 2
Well i couldnt do it with calculus but
a perpindicular line is the gradient * the perpindicular gradient = -1
we want to find a perpindicular line from 10,2 to a point of y = 2x-3
the gradient is 2 so 2*x=-1 x= -0.5
form a line y-2 = -0.5(x-10)
y=-0.5x+7
2x+3=-0.5x+7
x=4
y(4)=2*4-3
=5
so the point that intersects at the shortest possible point is (4,5)