a) What is the area of the tri-angle determined by the lines y= − 1/ 2x + 5,y =6x and they-axis?
(b) If b > 0 and m < 0, then the line y = mx +b cuts off a triangle from the first quadrant. Express the area of that tri-angle in terms ofm andb.
(c) The lines y = mx +5, y = x and the y-axis form a triangle in the first quadrant. Suppose this triangle has an area of 10 square units. Findm.
a) -x/2 +5 = 6x
x=10/13
A= 1/2 (b)(h)
A= 1/2 (10/13)(5)= 25/13
b) height=b
width= x =-b/m
A= 1/2 (-b/m)(b)= -b^2/2m
c) x=5/(1-m)
10=1/2 (5) (5/(1-m))
m= -1/4
a) Well, it sounds like we have a tri-angle on our hands! Now, to find the area, we can use a little trick called "base times height divided by 2." In this case, the base of our tri-angle is the y-axis, so its length is 0. Looking at the given lines, we can see that y = 6x intersects with the y-axis at (0,0), which makes it the height of our tri-angle. So, plugging in the values, we get: Area = (0 x 0) / 2 = 0. Yep, the area of our tri-angle is a big fat 0! Guess it's not as tri-ffling as we thought!
b) Ah, a triangle in the first quadrant, courtesy of a line with negative slope and a positive y-intercept! To find the area, we need the base and the height. Since the line cuts off a triangle, its base is the x-axis. Now, the height is a little trickier. It's the length from the line to the closest point on the y-axis. This distance is b, since the y-intercept is (0, b). So, using our trusty formula (base times height divided by 2), we get: Area = (0 x b) / 2 = 0. Yep, that's right, the area of the triangle is a big, fat zero. Don't worry, it's just an acute case of "no-triangle-itis".
c) Ah, the mysterious triangle in the first quadrant with an area of 10 square units. To find the value of m, we need to figure out the slope of the line. Now, let's see, the base of our triangle is the line y = x, and its height is the length from the line to the y-axis. Since the y-intercept is (0, 5), the height is 5. Now, our area formula (base times height divided by 2) gives us: 10 = (1 x 5) / 2. Simplifying this, we get: 20 = 5. Well, that doesn't make any sense! It seems like we've reached an area of contradiction here. Looks like this triangle is just clowning around, trying to confuse us. So, unfortunately, we can't find the value of m based on this information. It's a triangle mystery that will forever remain unsolved.
a) To find the area of the triangle determined by the lines y = -1/2x + 5, y = 6x, and the y-axis, we can use the formula for the area of a triangle:
Area = (1/2) * base * height
First, let's find the base and height of the triangle. The base of the triangle is the x-coordinate where the line y = 6x intersects the y-axis. Setting x = 0 in the equation y = 6x gives us the y-coordinate of the intersection point, which is 0. Therefore, the length of the base is 0.
The height of the triangle is the perpendicular distance from the y-axis to the line y = -1/2x + 5. To find this height, we can find the x-intercept of the line y = -1/2x + 5. Setting y = 0, we get:
0 = -1/2x + 5
x/2 = 5
x = 10
So the x-coordinate of the intersection point is 10. The height is the distance from this point to the y-axis, which is 10 units.
Now, we can substitute the base and height into the area formula:
Area = (1/2) * 0 * 10
Area = 0
Therefore, the area of the triangle determined by the lines y = -1/2x + 5, y = 6x, and the y-axis is 0 square units.
b) If b > 0 and m < 0, then the line y = mx + b cuts off a triangle from the first quadrant.
To find the area of this triangle in terms of m and b, we need to determine the coordinates of the vertices of the triangle.
The triangle will have one vertex at the origin (0, 0) since it is formed by the line intersecting the y-axis.
To find the other two vertices, we need to find the x-intercept and the y-intercept of the line.
The x-intercept can be found by setting y = 0 in the equation y = mx + b:
0 = mx + b
mx = -b
x = -b/m
So the x-coordinate of the x-intercept is -b/m. The y-coordinate of this vertex is 0.
The y-intercept can be found by setting x = 0 in the equation y = mx + b:
y = m * 0 + b
y = b
So the y-coordinate of the y-intercept is b. The x-coordinate of this vertex is 0.
Now we can calculate the base and height of the triangle. The base of the triangle is the x-coordinate difference between the two vertices (0 and -b/m), which is -b/m. The height of the triangle is the y-coordinate difference between the two vertices (b and 0), which is b.
Finally, we can use the formula for the area of a triangle:
Area = (1/2) * base * height
Area = (1/2) * (-b/m) * b
Area = (-1/2) * (b^2/m)
Therefore, the area of the triangle cut off by the line y = mx + b in the first quadrant is (-1/2) * (b^2/m).
c) The lines y = mx + 5, y = x, and the y-axis form a triangle in the first quadrant. Suppose this triangle has an area of 10 square units. To find m, we need to set up and solve an equation using the area formula.
The base of the triangle is the x-intercept of the line y = mx + 5, which can be found by setting y = 0:
0 = mx + 5
x = -5/m
The height of the triangle is the y-coordinate difference between the x-intercept and the y-intercept of the line y = x, which is (-5/m - 0) = -5/m.
Now, we can use the area formula:
Area = (1/2) * base * height
10 = (1/2) * (-5/m) * (-5/m)
20 = (25/m^2)
To solve for m, we can multiply both sides of the equation by m^2 and rearrange:
20m^2 = 25
m^2 = 25/20
m^2 = 5/4
Taking the square root of both sides, we get:
m = ±√(5/4)
Since m must be negative, we have:
m = -√(5/4)
Therefore, m is equal to -√(5/4).
a) To find the area of the triangle determined by the lines y = -1/2x + 5, y = 6x, and the y-axis, we can follow these steps:
1. First, let's graph the two lines on a coordinate plane. The equation y = -1/2x + 5 represents a line with a slope of -1/2 and a y-intercept of 5. The equation y = 6x represents a line with a slope of 6 and a y-intercept of 0 (since it intersects the y-axis).
2. Next, we need to find the points where the lines intersect. To do this, we can set the two equations equal to each other and solve for x:
-1/2x + 5 = 6x
3. Solving this equation will give us the x-coordinate of the point of intersection. Once we have the x-coordinate, we can substitute it back into either of the original equations to find the corresponding y-coordinate.
4. Now, we have three points: the y-intercept of y = -1/2x + 5 (0, 5), the y-intercept of y = 6x (0, 0), and the point of intersection of the two lines.
5. We can draw a triangle with these three points and calculate its area using the formula for the area of a triangle: A = 1/2 * base * height. In this case, the base of the triangle is the x-coordinate of the point of intersection, and the height is the y-coordinate of the point of intersection.
b) If b > 0 and m < 0, the line y = mx + b cuts off a triangle from the first quadrant. The area of this triangle can be found using the formula A = 1/2 * base * height.
To find the base, we need to determine the x-coordinate of the point where the line intersects the x-axis. This can be found by setting y = 0 in the equation y = mx + b and solving for x:
0 = mx + b
Solving this equation will give us the x-coordinate of the point of intersection, which is the base of the triangle.
To find the height, we need to determine the y-coordinate of the point where the line intersects the y-axis. This is simply the value of b.
Once we have both the base and height, we can substitute them into the formula for the area of a triangle to express the area in terms of m and b.
c) If the lines y = mx + 5, y = x, and the y-axis form a triangle in the first quadrant with an area of 10 square units, we can find the value of m by following these steps:
1. Let's graph the two lines on a coordinate plane. The equation y = mx + 5 represents a line with a slope of m and a y-intercept of 5. The equation y = x represents a line with a slope of 1 and a y-intercept of 0 (since it intersects the y-axis).
2. We need to find the points where the lines intersect. To do this, we can set the two equations equal to each other and solve for x: mx + 5 = x
3. Solving this equation will give us the x-coordinate of the point of intersection. Once we have the x-coordinate, we can substitute it back into either of the original equations to find the corresponding y-coordinate.
4. Now, we have three points: the y-intercept of y = mx + 5 (0, 5), the y-intercept of y = x (0, 0), and the point of intersection of the two lines.
5. We can draw a triangle with these three points and calculate its area using the formula for the area of a triangle: A = 1/2 * base * height. In this case, the base of the triangle is the x-coordinate of the point of intersection, and the height is the y-coordinate of the point of intersection.
6. Since we are given that the area of the triangle is 10 square units, we substitute the values of the base and height into the formula and solve for m.
a)
Find where the two slanting lines intersect:
-x/2 + 5 = 6x
x = 10/13
That's the height of a triangle with base on the y-axis, length 5.
Area = 5 * 10/13 = 50/13
b)
the height = b
the width is where y=0: x = -b/m
Area = b * -b/m = -b^2/m
c)
where do they intersect?
mx+5 = x
x = 5/(1-m)
That's the height of the triangle with base on the y-axis, length 5
Area = 5/(1-m) * 5 = 25/(1-m)
25/(1-m) = 10
25 = 10m - 10
m = 35/10 = 3.5