Question

triangle pqr has coordinates p(-4, -2), q(-1, -3), and r(-3, -5). what are the coordinates of the vertices of the image after a rotation of 270° counterclockwise about the origin?
a) p(2, -4), q(3, -1), and r(5, -3)
b) p(2, 4), q(3, 1), and r(5, 3)
c) p(-2, 4), q(-3, 1), and r(-5, 3)
d) p(4, 2), q(1, 3), and r(3, 5)

Explanation:

Step1: Recall the 270° counterclockwise rotation rule

The rule for a 270° counterclockwise rotation about the origin is \((x,y)\to(y,-x)\).

Step2: Apply the rule to point \(P(-4,-2)\)

For \(P(-4,-2)\), using the rule \((x,y)\to(y,-x)\), we substitute \(x = - 4\) and \(y=-2\). So \(P'=(-2,4)\) (since \(y=-2\) and \(-x = -(-4)=4\)).

Step3: Apply the rule to point \(Q(-1,-3)\)

For \(Q(-1,-3)\), substitute \(x=-1\) and \(y = - 3\) into the rule \((x,y)\to(y,-x)\). We get \(Q'=(-3,1)\) (since \(y=-3\) and \(-x=-(-1) = 1\)).

Step4: Apply the rule to point \(R(-3,-5)\)

For \(R(-3,-5)\), substitute \(x=-3\) and \(y=-5\) into the rule \((x,y)\to(y,-x)\). We get \(R'=(-5,3)\) (since \(y = - 5\) and \(-x=-(-3)=3\)).

Answer:

C) \(P'(-2,4)\), \(Q'(-3,1)\), and \(R'(-5,3)\)