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Answer:
- P₂ and x are both supplementary to N₁
- Q₁
- R = 90° -x
- ΔSMP ≅ ΔSMR ∴ PS ≅ SR
Step-by-step explanation:
1. Angles P₂ and N₁ are opposite angles of inscribed quadrilateral PMNQ, so are supplementary. Angles N₁ and N₂ form a linear pair, so are supplementary. Angles supplementary to the same angle (N₁) are congruent, hence P₂ = x ≅ N₂
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2. ΔPMQ is isosceles, so angle Q₁ is also congruent to x.
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3. In ΔPMQ, the sum of angles is 180°, so ...
M₁ +2x = 180°
Dividing by 2 gives ...
M₁/2 +x = 90°
Angle M₁ subtends arc PQ of circle M. Angle R inscribed in circle M subtends the same arc, so ...
R = (M₁/2)
R = 90° -x
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4. From the above, we know that angles N₂ and R are complementary (total 90°), so angle S₂ = 90°. Segment MS will only intersect chord PR at right angles at the midpoint of that chord.
Hence S is the midpoint of PR and PS = SR.
For
y=a(x-h)^2+k
vertex is (h,k)
given
y=1(x-2)^2-5
the vertex is (2,-5)
tha'ts quadrant 4
andswer is 2nd one
