Find the equation of the line through the point (3,4) which cuts from the first quadrant a triangle of minimum area.
Draw a line through (3,4) cutting the x-axis at (x,0) and the y-axis at (0,y)
Area = (1/2(x)(y)
using slopes,
(y-4)/-3 = +4/(x-3)
xy - 3y - 4x + 12 = 12
y(x-3) = 4x
y = 4x/(x-3)
then
A = (1/2)x(4x)/x-3)
= 2x^2/(x-3)
dA/dx = [(x-3)(4x) - 2x^2]/(x-3)^2 = 0 for a max/min of A
4x^2 - 12x - 2x^2 = 0
2x^2 - 12x = 0
2x(x-6)= 0
x = 0 or x = 6, and y = 24/3 = 8
using (0,8) and (3,4)
slope = (8-4)/-3
= -4/3
line has equation
y = (-4/3)x + 8
slope of line =
" using slopes,
(y-4)/-3 = +4/(x-3)
xy - 3y - 4x + 12 = 12 "
can you explain that part
nvr mind
Anyways thanks Reiny
Find the gradient of a line joining R(4,8) & S(5,-2).
To find the equation of the line through the point (3,4) which cuts from the first quadrant a triangle of minimum area, we need to find the slope of the line.
Since the line passes through the point (3,4), we can find its slope using the point-slope formula:
slope = (y2 - y1) / (x2 - x1),
where (x1, y1) = (3,4) and (x2, y2) are the coordinates of any other point on the line.
Let's assume the coordinates of another point on the line are (x, y).
The slope of the line can be calculated as:
slope = (y - 4) / (x - 3).
Now, we want the line to cut from the first quadrant a triangle of minimum area. For that, we want the triangle's base to lie on the x-axis (the line should intersect the x-axis), and its height to be as small as possible.
Since the base lies on the x-axis, the y-coordinate of the line where it intersects the x-axis will be 0. So, we can set y = 0 in the equation of the line:
0 = (0 - 4) / (x - 3).
Simplifying this equation, we get:
0 = -4 / (x - 3).
Solving for x, we have:
0 = -4.
However, this results in a division by zero, which is not possible. Hence, we cannot find a line that cuts from the first quadrant a triangle of minimum area.
Therefore, there is no equation of the line that satisfies the given condition.