A triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(0, 10), a
2 answers:
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Option B is right perpendicular bisectors of the sides of the triangle
<h3>What circumcircle?</h3>If we draw a circle through all the vertices of a triangle then the circle obtained is the circumcircle of the given triangle. We have in a triangle there are some properties inherent.
One is the perpendicular bisectors of all sides are concurrent. These three concur at a point called the circumcentre of the triangle.
When we draw a perpendicular bisector for a side AB, we get that all points on the perpendicular bisector are equidistant from A and B.
When we take the point of intersection of the perpendicular bisector of AB and BC, we have the intersecting point S(say) is equidistant from A and B and also from B and C
In short, this is equidistant from all 3 vertices and hence is the centre of the circumcircle of triangle ABC
Therefore Option B is right perpendicular bisectors of the sides of the triangle.
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The given inequality is y ≥ |x + 2| -3.
This inequality may be written two ways:
(a) y ≥ x + 2 - 3
or
y ≥ x - 1
(b) y ≥ -x -2 - 3
or
y ≥ -x - 5
A graph of the inequality is shown below. The shaded region satisfies the inequality.
Answer: A shaded region above a solid boundary line.
This inequality may be written two ways:
(a) y ≥ x + 2 - 3
or
y ≥ x - 1
(b) y ≥ -x -2 - 3
or
y ≥ -x - 5
A graph of the inequality is shown below. The shaded region satisfies the inequality.
Answer: A shaded region above a solid boundary line.