Question

quadrilateral (abcd) has vertices (a(-2,4)), (b(1,3)), (c(2,-3)), and (d(-3,-1)). quadrilateral (abcd) is rotated (90^{circ}) counterclockwise about (b). in which quadrant is the image of point (d)?

a) quadrant i

b) quadrant ii

c) quadrant iii

d) quadrant iv

Explanation:

Step1: Translate points

Translate the coordinate system so that point \(B(1,3)\) is the origin. The new coordinates of \(D(-3,-1)\) are \((-3 - 1,-1 - 3)=(-4,-4)\).

Step2: Apply rotation formula

The formula for a \(90^{\circ}\) counter - clockwise rotation about the origin \((x,y)\to(-y,x)\). For the point \((-4,-4)\), after rotation, it becomes \((4,-4)\).

Step3: Translate back

Translate the coordinate system back. Add the coordinates of \(B(1,3)\) to the rotated point \((4,-4)\). The new coordinates are \((4 + 1,-4+3)=(5,-1)\).

Answer:

D. Quadrant IV