The area of the triangle formed by x - and y - intercepts of the parabola y=0.5 (x-3) (x+k) is equal to 1.5 square units. Find all possible values of k.

K = - 1 or - 2

Step-by-step explanation:

Given the equation of parabola is, y = 0.5 (x-3) (x+k).

Finding y-intercept, put x=0 and solve for y : -

y = 0.5 (0-3) (0+k)

y = 0.5 (-3) (k)

y = - 1.5k

Finding x-intercepts, put y=0 and solve for x : -

0 = 0.5 (x-3) (x+k)

(x-3) (x+k) = 0

x = 3 OR x = - k

So, the coordinates of triangle ABC are:-

A (0, - 1.5k), B (3, 0), and C (-k, 0)

Assuming origin O (0,0), find the area of triangle ABC:-

Area = (1/2) * BC * OA

Area = (1/2) * (k+3) * (1.5k)

But given the area is 1.5 square units.

So, (1/2) * (-k-3) * (1.5k) = 1.5

(1/2) * (-k-3) * (k) = 1

k (-k-3) = 2

-k² - 3k - 2 = 0

k² + 3k + 2 = 0

k² + 2k + k + 2 = 0

(k+2) (k+1) = 0

So, k = - 1 or k = - 2

k = - 1,-2,-3,-5