The vertices of triangle qpr are q(–4,0), p(–6, 7), and r(–2, 1). / is rotated 90° counterclockwise about the origin to form q' p' r'. what are the coordinates of the vertices of q'p'r'?
a. q’(0, –4), p’(–7, –6), r’(–1, –2)
b. q’(0, –4), p’(7, 6), r’(1, –2)
c. q’(0, 4), p'(–7, 6), r’(–1, 2)
d. q’(4, 0), p’(6, 7), r’(2, –1)

 The correct option is

(A)  Q'(0,-4), P'(–7, -6) and R'(–1, -2).

Step-by-step explanation:  Given that the vertices of triangle QPR are Q(–4,0), P(–6, 7), and R(–2, 1). The triangle is rotated 90° counterclockwise about the origin to form Q' P' R'.

We are to find the co-ordinates of the vertices of triangle Q'P'R'.

We know that

if a point (x, y) is rotated 90° counterclockwise about the origin, then its co-ordinates becomes

(x, y)  ⇒  (-y, x).

So, the vertices of triangle QPR changes as follows :

Q(-4, 0) ⇒ Q' (0, -4),

P(-6,7)   ⇒  P'(-7, -6)

and

R(-2, 1)   ⇒  R'(-1, -2).

Thus, the co-ordinates of the vertices of triangle Q'P'R' are Q'(0,-4), P'(–7, -6) and R'(–1, -2).

Option (A) is CORRECT.

x=-2

infinite number of lines, so there is an infinite number of equations.

step-by-step explanation:

there are an infinite number of line that passes through the point (-2,5)

x=-2   is one line that passes through the point

y = 5 is another line that passes through the point

(these two lines are perpendicular to each other).

since the number of   numbers are infinite and the slope is the ratio of two numbers,(change in y over change in x) we can have an infinite number of slopes.     our only condition is that it passes through the point (-2,5)