Question

16
$\triangle stv$ has vertices $s(-3,-2)$, $t(-4,3)$ and $v(-2,3)$. if $(x,y)\to(x + 2,y - 3)$, what are the vertices of its image?
$s(-5,1)$, $t(-6,6)$, $v(-4,6)$
$s(-1,-5)$, $t(-2,0)$, $v(0,0)$
$s(3,2)$, $t(4,-3)$, $v(2,-3)$
$s(-1,-4)$, $t(-2,5)$, $v(1,6)$

Explanation:

Step1: Find the new \(x\)-coordinate for \(S\)

For point \(S(-3,-2)\), using the transformation \((x,y)\to(x + 2,y-3)\). The new \(x\)-coordinate is \(x=-3 + 2=-1\).

Step2: Find the new \(y\)-coordinate for \(S\)

The new \(y\)-coordinate is \(y=-2-3=-5\). So \(S'(-1,-5)\).

Step3: Find the new \(x\)-coordinate for \(T\)

For point \(T(-4,3)\), the new \(x\)-coordinate is \(x=-4 + 2=-2\).

Step4: Find the new \(y\)-coordinate for \(T\)

The new \(y\)-coordinate is \(y=3-3 = 0\). So \(T'(-2,0)\).

Step5: Find the new \(x\)-coordinate for \(V\)

For point \(V(-2,3)\), the new \(x\)-coordinate is \(x=-2+2 = 0\).

Step6: Find the new \(y\)-coordinate for \(V\)

The new \(y\)-coordinate is \(y=3-3=0\). So \(V'(0,0)\).

Answer:

\(S'(-1,-5)\), \(T'(-2,0)\), \(V'(0,0)\) (corresponding to the second option in the multiple - choice list)