A right triangle is formed in the first quadrant by the x- and y-axes and a line through the point (9,5).
Write the length L of the hypotenuse as a function of x.
I have no clue how to do this at all. Help.
A right triangle is formed in the first quadrant by the x- and y-axes and a line through the point (9,5).
Write the length L of the hypotenuse as a function of x.
I have no clue how to do this at all. Help.
Im imagining the end formula is going to be some sort of distance formula, how I get there and eliminate the Ys is beyond me.
I think I have to make some sort of imaginary line, have it connect with imaginary points, and find the distance of the imaginary points in a forumula for distance...
i seem to remember this is pretty early trigonometry but on account of being an english major i have forcefully forgot how to maths
the triangle is formed by the x axis and y axis. since it's a right triangle, the y intercept and x intercept are equal and the slope of the hypotenuse is -1.
use the point slope form of the equation of a line:
(y-y1)=m(x-x1)
(y-5)=-1(x-9)
y=-x+14
wait what?
just because its a right traingle doesn't mean the middle piece is sloped in any specific way. The x axis leg could extend far longer than the y axis leg, meaning the slope of the line connecting the two(hypotnuse) would be really small making the line really big.
Its not an equilateral right triangle.
there is no way to solve this without more information
hm im not sure immediately. i thought about constructing a function for y based on its limits but i dont think that'll work.
wait a second no it cant abifucwgofnqpegvfqwwd
im 99% sure im right.
The legth of it is based on the slope(the only unknown factor( so x) and a point on the line. I simplified a bunch of formulas into one easy formula, i can't believe i'm not correct.
Really someone explain how my formula doesn't work for every single slope value ever(slopes have to be negative btw)