Point Q is equidistant from AB and AC. What must be true about Q?
Chose from the following
A. It's on the perpendicular bisector of AB
B. It's on the perpendicular bisector of AC
C. BQ=CQ
D. It is on the angle bisector of ∠BAC.

D. It is on the angle bisector of ∠BAC

Step-by-step explanation:

Given that point Q is equidistant from AB and AC. This implies that AB and AC forms an angle BAC i.e <BAC. An equidistant point is a point that has the same distance from two different reference points.

Bisection is the process by which a line or angle is divided into two equal parts. A line that passes through the point of division is called a bisector, which is of equal distance to either of the reference points.

Thus by construction, a point that would be of equal distance to AB and AC is on an angle bisector of <BAC. So that point Q is on the angle bisector of <BAC.

answer is x<

18 units²

step-by-step explanation:

first let's find altitude bd using the distance formula

d = √((y₂ - y₁)² + (x₂ - x₁)²)

bd = √((0 - -3)² + (0 - 3)²)

bd = √18

given the placement of points a and c, we can denote that point d bisects segment ac.

we will call ac the base.

using the distance formula

ac = √((0 - 6)² + (-6 - 0)²)

ac = √72

knowing our base and our height, we can plug these into our triangle area formula

area = (b * h)/2

area = (√18 * √72) / 2

area = (√1296) / 2

area = 36 / 2

area = 18 units²