a right triangle is formed in the first quadrant by the x- and y- axes and a line through the point (1,2). write the length of the hypotenuse as a function of x. find the vertices of the triangle such that its area is a minimum.
Lee bro you were 8 years late lmaoo
let P(x,0) be the x-intercept and Q(0,y) be the y-intercept
label (1,2) as A(1,2)
the slope AP = slope AQ
-2/(x-1) = (2-y)/1
2x - xy - 2 + y = -2
y(1-x) = -2x
y = -2x/(1-x) or 2x/(x-1)
H^2 = x^2 + y^2
= x^2 + ( (2x/(x-1) )^2 --- simplify if needed
(don't know why we are finding the hypotenuse ? )
Area of triangle OPQ
= (1/2)base x height
= (1/2)xy
= (1/2)x(2x/(x-1))
= (1/2) (2x^2/(x-1) )
dA/dx = (1/2) [ (x-1)(4x) - 2x^2(1) ]/(x-1)^2
= (1/2) [ 4x^2 - 4x - 2x^2 ]/(x-1)^2
= 0 for a max of A
2x^2 - 4x = 0
x(2x-4) = 0
x=0 ---- > makes no sense, look at diagram
or
x = 2
then y = 2(2)/(2-1) = 4
the triangle has (2,) and (0, 4) as its vertices for a maximum area
sticky keyboard, last line should obviously say
the triangle has (2,0) and (0, 4) as its vertices for a maximum area
Hi sorry for the late response but this question asked for the minimum but you did the maximum. Was this a typo on your part or would the answer be different for the minimum?
Oh boy, a triangle question! Let's put on our mathematical clown hats and have some fun!
To find the length of the hypotenuse as a function of x, we need to determine the equation of the line that passes through the point (1,2). Let's call the hypotenuse length H(x).
Since the line passes through (1,2), we can use the point-slope form of a line equation, which is y - y1 = m(x - x1). We can substitute (1,2) as the point (x1, y1) to find the equation of the line passing through it.
Using the point-slope form, we have:
y - 2 = m(x - 1)
Now, since the line forms a right triangle with the x-axis and y-axis, the slope of the line will be the negative reciprocal of the slope of the axes, which is 1. So, m = -1.
Substituting -1 for m in the equation, we have:
y - 2 = -1(x - 1)
y - 2 = -x + 1
y = -x + 3
Now, to find the length of the hypotenuse, we need to find the coordinates where the line intersects the x and y-axes. The y-intercept occurs when x = 0, so we substitute x = 0 into the equation of the line:
y = -(0) + 3
y = 3
Therefore, one vertex of the triangle is (0,3).
The x-intercept occurs when y = 0, so we substitute y = 0 into the equation of the line:
0 = -x + 3
x = 3
Therefore, another vertex of the triangle is (3,0).
Now, we have the vertices (1,2), (0,3), and (3,0).
To find the length of the hypotenuse H(x), we can use the Pythagorean theorem. The hypotenuse is the longest side of a right triangle and is opposite the right angle.
The distance formula between two points (x1, y1) and (x2, y2) is √((x2 - x1)^2 + (y2 - y1)^2).
Applying this formula to find the length of the hypotenuse between the vertices (1,2) and (3,0), we have:
H(x) = √((3 - 1)^2 + (0 - 2)^2)
H(x) = √(2^2 + (-2)^2)
H(x) = √(4 + 4)
H(x) = √8
H(x) = 2√2
So, the length of the hypotenuse as a function of x is H(x) = 2√2.
As for finding the vertices of the triangle where its area is a minimum, that would require more information or constraints. Without additional information, we can't determine the vertices with the minimum area.
Hope this clowning around with math was entertaining! Let me know if there's anything else I can humorously assist you with!
To find the length of the hypotenuse of the right triangle as a function of x, we need to determine the equation of the line passing through the point (1,2).
Let's denote the length of the hypotenuse as h, and the other two sides of the right triangle as a and b. Using the Pythagorean theorem, we know that h^2 = a^2 + b^2.
Now, to find the equation of the line passing through (1,2), we need its slope and y-intercept. We can obtain the slope using the point-slope form of a linear equation: y - y1 = m(x - x1).
Let's call the slope of the line m. We have the point (x1, y1) = (1,2). Plugging these values into the equation, we get y - 2 = m(x - 1).
To find the slope, we need another point on the line. Since the line passes through the origin (0,0), we can substitute these coordinates into the equation: 0 - 2 = m(0 - 1).
Simplifying, we get -2 = -m. Dividing both sides by -1, we find that m = 2.
Now that we have the slope, we can rewrite the equation as y - 2 = 2(x - 1). Expanding and rearranging, we get y = 2x.
To find the length of the hypotenuse as a function of x, we need to determine the coordinates where the line y = 2x intersects with the x and y-axes.
To find the x-intercept, we substitute y = 0 into the equation y = 2x: 0 = 2x. Solving for x, we find x = 0.
To find the y-intercept, we substitute x = 0 into the equation y = 2x: y = 2(0). Simplifying, y = 0.
Therefore, the coordinates of the vertices of the triangle are (0,0), (1,2), and (0,2).
The area of a right triangle is given by the formula: A = (1/2) * base * height. In this case, the base is given by the x-coordinate of the point (1,2), which is 1, and the height is given by the y-coordinate of the point (0,2), which is 2.
Plugging these values into the formula, we get A = (1/2) * 1 * 2 = 1.
Hence, the area of the triangle is minimum when the length of the hypotenuse (h) is a constant value, and the coordinates of the triangle's vertices are (0,0), (1,2), and (0,2).