Question

question 15
1 pts
if a point in quadrant ii is reflected in the y-axis, its image will lie in quadrant ____.
options: iii, i, on the y-axis, ii
question 16
1 pts
△abc, with vertices a(2,0), b(3,2) and c(3,−1), is rotated 90° counterclockwise about the origin. what are the coordinates of the vertices of the rotated triangle?
options: a′(0,2), b′(−2,−3), c′(1,−3); a′(0,−2), b′(−2,3), c′(1,3); a′(0,2), b′(−2,3), c′(1,3); a′(0,−2), b′(−2,−3), c′(1,−3)
question 17
1 pts
what is the image of (5,1) under a counterclockwise rotation of 90°?

Explanation:

Question 15

A point in Quadrant II has a negative x - coordinate and a positive y - coordinate. When reflecting over the y - axis, the rule is \((x,y)\to(-x,y)\). So a point \((-a,b)\) (where \(a>0,b > 0\)) in Quadrant II will become \((a,b)\) after reflection over the y - axis, which is in Quadrant I.

Step1: Recall the rotation rule

The rule for a \(90^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\).

Step2: Rotate point \(A(2,0)\)

For point \(A(2,0)\), applying the rule \((x,y)\to(-y,x)\), we get \(A'=( - 0,2)=(0,2)\).

Step3: Rotate point \(B(3,2)\)

For point \(B(3,2)\), applying the rule \((x,y)\to(-y,x)\), we have \(x = 3,y = 2\), so \(B'=(-2,3)\).

Step4: Rotate point \(C(3,-1)\)

For point \(C(3,-1)\), applying the rule \((x,y)\to(-y,x)\), we have \(x = 3,y=-1\), so \(C'=(1,3)\).

Step1: Recall the rotation rule

The rule for a \(90^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\).

Step2: Apply the rule to the point \((5,1)\)

For the point \((5,1)\), let \(x = 5\) and \(y = 1\). Using the rule \((x,y)\to(-y,x)\), we get the new point as \((-1,5)\).

Answer:

B. I

Question 16