Find the area of triangle whose vertices are (- 8,4 )(- 6,6) and (- 3,9)

Area of the triangle = 0

Step-by-step explanation:

We are given the vertices of a triangle as: (- 8,4 ), (- 6,6), (- 3,9)

The formula to find the Area of the triangle =

1/2[ x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

Where :

(x₁, y₁) =  (- 8,4 )

(x₂, y₂) = (- 6,6)

(x₃, y₃) =  (- 3,9)

Area of the triangle = 1/2[-8(6 - 9) + -6(9 - 4) + -3(4 - 6)]

= 1/2[ (-8 × -3) +( -6 × 5) +( -3× -2)]

= 1/2[ 24 - 30 + 6)

= 1/2[ 24 + 6 - 30]

= 1/2 [30 - 30]

=1/2[ 0 ]

= 0

Therefore, the area of triangle whose vertices are (- 8,4 ), (- 6,6) and (- 3,9) is ZERO( = 0 )

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

yes

step-by-step explanation: