Question

a triangle has vertices at r(1, 1), s(-2, -4), and t(-3, -3). the triangle is transformed according to the rule ( r_{0, 270^circ} ). what are the coordinates of ( s )? ( \bigcirc (-4, 2) ) ( \bigcirc (-2, 4) ) ( \bigcirc (2, 4) ) ( \bigcirc (4, -2) )

Explanation:

Step1: Recall 270° rotation rule

The rule for a 270° rotation about the origin (\(R_{O,270^\circ}\)) is \((x,y)\to(y, -x)\).

Step2: Apply rule to point S(-2, -4)

For point \(S(-2, -4)\), \(x = -2\) and \(y = -4\). Using the rule \((x,y)\to(y, -x)\), we substitute \(x\) and \(y\):
New \(x\)-coordinate: \(y=-4\)
New \(y\)-coordinate: \(-x=-(-2) = 2\)
Wait, no, wait. Wait, the correct rule for 270° counterclockwise rotation about the origin is \((x,y)\to(y, -x)\)? Wait, no, actually, the rule for 270° counterclockwise (or 90° clockwise) rotation about the origin is \((x,y)\to(y, -x)\)? Wait, no, let's correct. The standard rotation rules:

• 90° counterclockwise: \((x,y)\to(-y,x)\)
• 180° counterclockwise: \((x,y)\to(-x,-y)\)
• 270° counterclockwise (or 90° clockwise): \((x,y)\to(y, -x)\)

Wait, let's take a point (a,b). Rotating 270° counterclockwise around the origin: the formula is \((x,y)\to(y, -x)\). So for \(S(-2, -4)\), \(x=-2\), \(y = -4\). Then new \(x\) is \(y=-4\), new \(y\) is \(-x=-(-2)=2\)? Wait, but that would be (-4, 2)? Wait, no, wait, maybe I mixed up. Wait, let's check with a simple point. Let's take (1,0). Rotating 270° counterclockwise: it should go to (0, -1)? Wait, no, (1,0) rotated 90° counterclockwise is (0,1), 180° is (-1,0), 270° is (0, -1). Using the formula \((x,y)\to(y, -x)\): (1,0) becomes (0, -1), which is correct. So for (x,y)=(1,0), y=0, -x=-1, so (0, -1). Correct. So for S(-2, -4), x=-2, y=-4. Then new x is y=-4, new y is -x=-(-2)=2. So S' is (-4, 2)? Wait, but let's check the options. Option A is (-4,2). Wait, but let's re - derive the rotation rule.

Alternative way: The rotation of a point \((x,y)\) by 270° counterclockwise about the origin can be derived from the rotation matrix. The rotation matrix for \(\theta\) degrees counterclockwise is \(

$$\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$

\). For \(\theta = 270^\circ\), \(\cos(270^\circ)=0\), \(\sin(270^\circ)= - 1\). So the matrix is \(

$$\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

\). Multiplying this matrix with the vector \(

$$\begin{pmatrix}x\\y\end{pmatrix}$$

\) gives \(

$$\begin{pmatrix}y\\-x\end{pmatrix}$$

\), which is the same as the rule \((x,y)\to(y, -x)\).

So for point \(S(-2,-4)\), applying the rule \((x,y)\to(y, -x)\):
\(x=-2\), \(y = -4\)
New \(x\) - coordinate: \(y=-4\)
New \(y\) - coordinate: \(-x=-(-2)=2\)
So the coordinates of \(S'\) are \((-4,2)\)

Answer:

\((-4, 2)\) (corresponding to the first option: \(\boldsymbol{(-4, 2)}\))