Answer:
- 3) y = (7/2)x -12
- 4) y = 3x -5
- 7) y = (1/3)x +4/3
- 8) y = (-4/3)x +8/3
Step-by-step explanation:
In every case, you can ...
- replace any constant in the equation by 0
- for point (h, k), replace x with (x-h) and y with (y-k), then simplify
- solve for y (add the opposite of the y-term; divide by the y-coefficient)
3) -7(x-4) +2(y-2) = 0 ⇒ -7x +2y +24 = 0
... y = (7/2)x -12
4) (y+2) = 3(x-1)
... y = 3x -5
7) -(x+4) +3(y-0) = 0 ⇒ -x +3y -4 = 0
... y = (1/3)x +4/3
8) 4(x-2) +3(y -0) = 0 ⇒ 4x +3y -8 = 0
... y = (-4/3)x +8/3
Are you asking what the answer is because you are not asking us anything. I would really like to help you out but I need to know what the question is first.
Note: Consider we need to find the vertices of the triangle A'B'C'
Given:
Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.
Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).
To find:
The vertices of the triangle A'B'C'.
Solution:
If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then
Using this rule, we get
Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).
The answer to your problem is a<-2
