Question

question 9
1 pts
if p(3, -4) is reflected on the point (3, 0), what are the coordinates of p, the image of p?
(3, 4)
(-3, -4)
(4, 3)
(3, -4)
question 10
1 pts
what is the image of (-4, 1) after a rotation of 180° clockwise?
(1, 4)
(1, -4)
(4, -1)
(-1, -4)
question 11
1 pts
what is the image of (4, 2) under a counterclockwise rotation of 90°?

Explanation:

Question 9

Step1: Recall reflection over a point

For a point \( P(x,y) \) reflected over the point \( (a,b) \), the midpoint formula is used. The midpoint \( M \) between \( P(x,y) \) and \( P'(x',y') \) is \( (a,b) \), so \( \frac{x + x'}{2}=a \) and \( \frac{y + y'}{2}=b \).

Step2: Apply midpoint formula

Given \( P(3,-4) \) and midpoint \( (3,0) \). For the x - coordinate: \( \frac{3 + x'}{2}=3 \), multiply both sides by 2: \( 3 + x' = 6 \), so \( x'=3 \). For the y - coordinate: \( \frac{-4 + y'}{2}=0 \), multiply both sides by 2: \( -4 + y' = 0 \), so \( y' = 4 \). Thus, \( P'=(3,4) \).

Step1: Recall \( 180^{\circ} \) rotation rule

The rule for a \( 180^{\circ} \) clockwise (or counter - clockwise) rotation about the origin is \( (x,y)\to(-x,-y) \).

Step2: Apply the rotation rule

For the point \( (-4,1) \), applying the rule \( (x,y)\to(-x,-y) \), we get \( x'=-(-4)=4 \) and \( y'=-1 \). So the image is \( (4,-1) \).

Step1: Recall \( 90^{\circ} \) counter - clockwise rotation rule

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

Step2: Apply the rotation rule

For the point \( (4,2) \), using the rule \( (x,y)\to(-y,x) \), we have \( x'=-2 \) and \( y' = 4 \). Wait, no, wait: the correct rule is \( (x,y)\to(-y,x) \). So if \( x = 4 \) and \( y = 2 \), then \( x'=-y=-2 \)? No, wait, I made a mistake. Wait, the correct rule for \( 90^{\circ} \) counter - clockwise rotation is \( (x,y)\to(-y,x) \). Wait, no, let's re - derive it. A \( 90^{\circ} \) counter - clockwise rotation: if we have a vector \( \vec{v}=(x,y) \), after a \( 90^{\circ} \) counter - clockwise rotation, the new vector \( \vec{v}'=(-y,x) \). So for \( (4,2) \), \( x = 4 \), \( y = 2 \), so \( x'=-y=-2 \)? No, wait, no, let's take a simple example. Let's take the point \( (1,0) \), a \( 90^{\circ} \) counter - clockwise rotation gives \( (0,1) \), which fits \( (x,y)\to(-y,x) \) (since \( x = 1,y = 0 \), \( -y = 0,x = 1 \)). For \( (0,1) \), a \( 90^{\circ} \) counter - clockwise rotation gives \( (-1,0) \), which is \( (-y,x) \) ( \( x = 0,y = 1 \), \( -y=-1,x = 0 \)). For \( (4,2) \), \( x = 4 \), \( y = 2 \), so \( -y=-2 \), \( x = 4 \)? No, wait, I messed up. Wait, the correct rule is \( (x,y)\to(-y,x) \). Wait, no, actually, the standard rule for \( 90^{\circ} \) counter - clockwise rotation is \( (x,y)\to(-y,x) \). So for \( (4,2) \), \( x = 4 \), \( y = 2 \), so the new point is \( (-2,4) \). Wait, let's check with a graph. If we have a point \( (4,2) \), moving 90 degrees counter - clockwise: the x - coordinate becomes the negative of the y - coordinate, and the y - coordinate becomes the x - coordinate. So \( (4,2)\to(-2,4) \).

Answer:

A. \((3, 4)\)

Question 10