Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

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

Step-by-step explanation: