If the x-intercept is at x=h, then we must have h>2 to fit the conditions.
Now we have two points, and our line is
(y-3)/(x-2) = (0-3)/(h-2)
y = 3/(2-h) * (x-2)
at x=0, y = 6/(h-2)
the area is thus 1/2 xy = 1/2 h(6/(h-2)) = 3h/(h-2)
A right triangle is formed in the first quadrant by the x- and y-axes and a line through the point (2,3). Draw a figure that illustrates the problem.
What is the area of the triangle A as a function of the x-intercept of the line?
1 answer