Determine whether the three points are collinear. (0, -10), (-3, -15), (2,- 6)

Three points are collinear if their determinant is equal to 0

So,

| 0  -10 1 |

| -3 -15 1 |

| 2   -6 1 |

We must put 1 at the third column since we don't have z in these points, and 1 is a neutral factor.

Applying sarrus

( 0 . -15 . 1 + -10 . 1 . 2 + 1 . -3 . -6) - (1 . -15 . 2 + 0 . 1 . -6 + -10 . -3 . 1)

(0 - 20 + 18) - (-30 + 0 + 30)

(-2) - (0)

-2

Our determinant is different than 0, so these points are not collinear.

x = 12

y = 15

step-by-step explanation:

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i