The image of the positive ratio function y = x and the image of the inverse ratio function y = x (K ≠ 0) intersect with point a at the first term limit, and make a vertical line of X axis through point a, with the perpendicular foot M

The image of the positive ratio function y = x and the image of the inverse ratio function y = x (K ≠ 0) intersect with point a at the first term limit, and make a vertical line of X axis through point a, with the perpendicular foot M

A (root K, root K), m (root K, 0)

We know that the positive scale function passes through the point (- 2,5), passes through a point a on the image and makes a vertical line of the Y axis. The coordinates of the perpendicular foot B are (0, - 3)
(1) Coordinates of point a
(2) Area of triangle AOB

y=kx
Over (- 2,5)
5=-2k
k=-5/2
y=-5x/2
AB is perpendicular to y, so the ordinates are equal
So a (a, - 3)
So - 3 = - 5A / 2
a=6/5
A(6/5,-3)
AB=|0-6/5|=6/5
OB=|-3|=3
So area = (6 / 5) * 3 * 2 = 9 / 5

If there is a point a on the positive scale function y = 3x, a vertical line passing through the point a and the coordinate of the perpendicular foot B is (- 2,0), then the area of △ AOB is?

Because the coordinate of perpendicular B is (- 2,0)
The base of ∧ AOB is ∧ - 2 ∧ = 2
And ∵ y = 3x
∴y=3（-2）=-6
The height of ∧ is ∧ - 6 ∧ = 6
That is s △ AOB = (AB · OB) △ 2 = 2 × 6 △ 2 = 6