Find a value of m so that the triangle enclosed by the line x+y=m and the coordinate axes will have an area of 200 square units.
The triangle enclosed by the line x+y=m and the coordinate axes is a right-angled triangle with legs parallel to the coordinate axes. Let's call the length of the horizontal leg of the triangle x, and the length of the vertical leg y. Then, the area of the triangle is:
A = (1/2)xy
We can use the equation of the line to express x and y in terms of m:
x + y = m
x = m - y
y = m - x
Substituting these expressions into the equation for the area, we get:
A = (1/2)(m - y)y
A = (1/2)(m - x)x
Combining these two expressions and simplifying, we get:
A = (1/2)(m - x)x = (1/2)(m - y)y
A = (1/2)(m^2 - xy)
2A = m^2 - xy
We know that the area of the triangle is 200 square units, so we can substitute that into the equation and solve for m:
2A = m^2 - xy
2(200) = m^2 - xy
400 = m^2 - xy
We need to find values of x and y that satisfy the equation x + y = m and the equation 400 = m^2 - xy. We can substitute x = m - y into the second equation to get:
400 = m^2 - x(m - x)
400 = m^2 - mx + x^2
Rearranging and factoring, we get:
x^2 - mx + (m^2 - 400) = 0
This is a quadratic equation in x. We can use the quadratic formula to solve for x:
x = [m ± sqrt(m^2 - 4(m^2 - 400))] / 2
x = [m ± sqrt(1600 - 3m^2)] / 2
For the triangle to be a right-angled triangle, one of the legs must have length 10 times larger than the other leg. Let's assume that the horizontal leg is 10 times larger than the vertical leg, so x = 10y. Substituting this into the equation above, we get:
10y = [m ± sqrt(1600 - 3m^2)] / 2
20y = m ± sqrt(1600 - 3m^2)
We know that y is a positive number, so we can take the positive root:
20y = m + sqrt(1600 - 3m^2)
Squaring both sides, we get:
400y^2 = m^2 + 1600 - 2m sqrt(1600 - 3m^2)
Rearranging, we get:
2m sqrt(1600 - 3m^2) = m^2 - 400y^2 + 1600
Squaring both sides again, we get:
4m^2 (1600 - 3m^2) = (m^2 - 400y^2 + 1600)^2
Expanding and simplifying, we get:
-3m^4 + 400m^2 - 1600y^2 = 0
This is a quadratic equation in m^2. We can use the quadratic formula to solve for m^2:
m^2 = [-(400) ± sqrt(400^2 - 4(-3)(-1600y^2))] / (-6)
m^2 = [200 ± 10sqrt(200 + 3y^2)] / 3
=[200 ± 388]/3
We need the positive root, m^2 = (200 + 388)/3 = 196. Therefore, m = 14.
+20 To find the value of m such that the triangle enclosed by the line x+y=m and the coordinate axes has an area of 200 square units, we can use the formula for the area of a triangle:
Area = 1/2 * base * height
In this case, the base of the triangle is the distance between the x-intercept and the point where the line x+y=m intersects the y-axis. The height of the triangle is the distance between the y-intercept and the point where the line x+y=m intersects the x-axis.
To find these points, we can set x=0 and y=0 in the equation x+y=m, and solve for the corresponding variables:
When x=0, y=m, so the line intersects the y-axis at the point (0,m).
When y=0, x=m, so the line intersects the x-axis at the point (m,0).
The distance between the x-intercept and the point (0,m) is m, and the distance between the y-intercept and the point (m,0) is m. Therefore, the base and height of the triangle are both equal to m.
Plugging in the values of base and height into the formula for the area, we get:
Area = 1/2 * m * m = 1/2 * m^2
We want the area to be 200 square units, so we can set up an equation:
1/2 * m^2 = 200
Solving for m, we get:
m^2 = 400
m = ±20
Since we are looking for a positive value of m, the answer is:
m = 20
Therefore, the value of m such that the triangle enclosed by the line x+y=20 and the coordinate axes has an area of 200 square units is 20.
