Find an equation of the line through the point that cuts off the least area from the first quadrant.
Please enter your answer in the slope-intercept form.
What were the coordinates of the point?
(4,16)... sorry about that.. thanks!
To find an equation of the line that cuts off the least area from the first quadrant and passes through the point (4, 16), we need to consider that the line should be tangent to the axes in order to minimize the area.
Since the line is tangent to both the x-axis and the y-axis, we can infer that it should be perpendicular to the axes. Let's assume this line has a slope m.
For the line to be tangent to the x-axis, the perpendicular line should pass through the point (4, 0). Using the point-slope form of a line, we have:
y - 0 = m(x - 4)
y = mx - 4m .......(1)
For the line to be tangent to the y-axis, the perpendicular line should pass through the point (0, 16). Using the point-slope form again, we have:
16 - 0 = m(0 - 4)
16 = -4m
m = -4
Substituting m = -4 back into equation (1), we have:
y = -4x - 4(-4)
y = -4x + 16
Therefore, the equation of the line that cuts off the least area from the first quadrant and passes through the point (4, 16) is y = -4x + 16 in slope-intercept form.