Question

jayson graphs logo b in the coordinate plane to find the points of each triangle that intersect with the circle. the graph is centered at (0,0) and the unit size of the coordinate plane is 1 foot.
which statement is true about the points on the circle?
a the triangle in quadrant i intersects the circle at point (1,1). when the triangle is reflected over the x - axis, it intersects the circle at point (1, - 1).
b the triangle in quadrant i intersects the circle at point (\\(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\\)). when the triangle is reflected over the x - axis, it intersects the circle at point (\\(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\\)).
c the triangle in quadrant i intersects the circle at point (1,1). when the triangle is reflected over the y - axis, it intersects the circle at point (1, - 1).
d the triangle in quadrant i intersects the circle at point (\\(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\\)). when the triangle is reflected over the y - axis, it intersects the circle at point (\\(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\\)).

Explanation:

Step1: Recall the equation of the unit circle

The equation of a unit circle centered at \((0,0)\) is \(x^{2}+y^{2}=1\). For a point \((x,y)\) to be on the unit circle, it must satisfy this equation.

• For point \((1,1)\): \(1^{2}+1^{2}=1 + 1=2

eq1\). So points like \((1,1)\) are not on the unit circle.

• For point \((x = \frac{\sqrt{2}}{2},y=\frac{\sqrt{2}}{2})\): \((\frac{\sqrt{2}}{2})^{2}+(\frac{\sqrt{2}}{2})^{2}=\frac{2}{4}+\frac{2}{4}=\frac{1}{2}+\frac{1}{2}=1\)

Step2: Recall the rule for reflection over the \(x -\)axis

When a point \((x,y)\) is reflected over the \(x -\)axis, the new point is \((x,-y)\).

Step3: Analyze each option

• Option A: Since \((1,1)\) is not on the unit circle (\(1^{2}+1^{2}=2

eq1\)), this option is incorrect.

• Option B:
• The point \((\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\) is on the unit circle (\((\frac{\sqrt{2}}{2})^{2}+(\frac{\sqrt{2}}{2})^{2}=1\)).
• When reflected over the \(x -\)axis, using the rule \((x,y)\to(x, - y)\), the new point is \((\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})\)
• Option C: Since \((1,1)\) is not on the unit circle (\(1^{2}+1^{2}=2

eq1\)), this option is incorrect.

• Option D: When a point \((x,y)\) is reflected over the \(y -\)axis, the rule is \((x,y)\to(-x,y)\). If the original point is \((\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\), the reflected point over the \(y -\)axis is \((-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})

eq(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})\)

Answer:

B. The triangle in Quadrant I intersects the circle at point \((\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\). When the triangle is reflected over the \(x -\)axis, it intersects the circle at point \((\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})\)