so, two of its vertices are
(-h,-k) and (1+h,-k)
Thus, the length of the base is (2h+1)
That means the altitude is (1/2 + h)√3, making its coordinates (3/2, k+(1/2 + h)√3)
See whether you can work out the relationship between h and k, and thus the equation for the path followed by C.
An equilateral triangle ⌂ABC
has two of its vertices below the x-axis,
has the third vertex C above the x-axis, and
contains the points A=(0,0)and B=(1,0) on its sides.
How long is the path traced out by all possible points C, to two decimal places?
3 answers
Nah - that's just too complicated.
Take a look at an equilateral triangle, and slice off the top of it. (That's the piece of the x-axis between the sides of the triangle.) That small cap of the big triangle is also an equilateral triangle, and always has the same base length.
So, point C stays the same, no matter how big the triangle!
Take a look at an equilateral triangle, and slice off the top of it. (That's the piece of the x-axis between the sides of the triangle.) That small cap of the big triangle is also an equilateral triangle, and always has the same base length.
So, point C stays the same, no matter how big the triangle!
OK - I may have misread it again, assuming that the base of the triangle is parallel to the x-axis.
This is much trickier than it looks, unless the size of the triangle is somehow determined.
This is much trickier than it looks, unless the size of the triangle is somehow determined.