B and c, anything with an x^2 term is a quadratic.
Answer:
none of them
Step-by-step explanation:
Two lines are perpendicular when satisfy the next equation: m1*m2 = -1, where m1 and m2 are the slopes o the lines.
line 1:
y – 1 = (x+2)
y = x + 3
slope of line 1 = 1
line 2:
y + 2 = –3(x – 4)
y + 2 =
-3*x + 12
y = -3*x + 10
slope of line 2 = -3
m1*m2 = 1*(-3
) = -3
They are not perpendicular
line 3:
y − 5 = 3(x + 11)
y − 5 = 3*x + 33
y = 3*x + 38
slope of line 3 = 3
m1*m3 = 1*3 = 3
They are not perpendicular
line 4:
y = -3x –
slope of line 4 = -3
m1*m4 = 1*(-3
) = -3
They are not perpendicular
line 5:
y = x – 2
slope of line 5 = 1
m1*m5 = 1*1 = 1
They are not perpendicular
line 6:
3x + y = 7
y = -3x + 7
slope of line 6 = -3
m1*m6 = 1*(-3
) = -3
They are not perpendicular
<u>Given</u>:
The coordinates of the points A, B and C are (3,4), (4,3) and (2,1)
The points are rotated 90° about the origin.
We need to determine the coordinates of the point C'.
<u>Coordinates of the point C':</u>
The general rule to rotate the point 90° about the origin is given by

Substituting the coordinates of the point C in the above formula, we get;

Therefore, the coordinates of the point C' is (-1,2)