Step-by-step explanation:
On a polar plane, the shape will be a line that goes through the pi/2- pi quadrant , and 3pi/2- 2pi quadrant
On a Cartesian or rectangular plane, the shape will be a line as well passing through 2nd and 4th quadrant
You don't need to understand the construction or why it works. You only need to accept the fact that it does. You can figure out the answers to this question by looking at the picture.
RT is tangent to circle Q -- TRUE. That is the point of the construction.
QT is a radius of circle Q -- TRUE. Q is the center and T is on the circle. A line segment from the center to a point on the circle is a radius.
m∠QSR = 90° -- FALSE. Those points lie on the same line. The measure of the angle is 180°.
QS = QT -- FALSE. S lies inside circle Q, so is closer to the center than T, which lies on the circle. (For some choice of point R, S might lie on the circle, but because this statement is not always true, it must be considered false.)
ΔRTQ is a right triangle -- TRUE. A tangent line is always perpendicular to the radius to the point of tangency. The construction succeeds because RTQ is inscribed in semicircle RTQ (centered at S). Such a triangle is always a right triangle.
The answer to this question will be C
