An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(7 ,5 )# to #(8 ,2 )# and the triangle's area is #27 #, what are the possible coordinates of the triangle's third corner?
The coordinates are
The area of the triangle is
The equation of the altitude is
The circle with equation
The intersection of this circle with the altitude will give the third corner.
We solve this quadratic equation
graph{(y1/3x1)((x7.5)^2+(y3.5)^2291.6)((x7)^2+(y5)^20.05)((x8)^2+(y2)^20.05)(y5+3(x7))=0 [12, 28, 10, 10]}
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To find the possible coordinates of the triangle's third corner, we can use the formula for the area of a triangle, which is given by:
[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ]
Since the triangle is isosceles, we can consider side ( A ) as the base. Then, the length of side ( A ) can be found using the distance formula between the given points ((7, 5)) and ((8, 2)).
Once we have the length of side ( A ) and the area of the triangle, we can find the height of the triangle. With the height, we can then find the possible coordinates of the third corner by considering the symmetry of the isosceles triangle.
Let's calculate:

Calculate the length of side ( A ): [ A = \sqrt{(8  7)^2 + (2  5)^2} = \sqrt{1 + 9} = \sqrt{10} ]

Use the area formula to find the height: [ \text{Area} = \frac{1}{2} \times A \times \text{height} ] [ 27 = \frac{1}{2} \times \sqrt{10} \times \text{height} ] [ \text{height} = \frac{2 \times 27}{\sqrt{10}} = \frac{54}{\sqrt{10}} ]

With the height, we can find the coordinates of the third corner. Since the triangle is isosceles, the third corner will be at the same distance from the midpoint of side ( A ) along the perpendicular bisector of side ( A ).
Let ( M ) be the midpoint of side ( A ). The coordinates of ( M ) can be found as the average of the coordinates of the given points:
[ M = \left(\frac{7 + 8}{2}, \frac{5 + 2}{2}\right) = \left(\frac{15}{2}, \frac{7}{2}\right) ]
Now, we need to find a point ( P ) that is ( \frac{54}{\sqrt{10}} ) units away from ( M ) along the perpendicular bisector of ( A ).
Since the perpendicular bisector of a line segment passes through its midpoint, we only need to find a point on the perpendicular line that is ( \frac{54}{\sqrt{10}} ) units away from ( M ).
The equation of the line passing through ( M ) perpendicular to side ( A ) is given by: [ x  \frac{15}{2} = \frac{3}{1} \times (y  \frac{7}{2}) ]
We solve this equation with the condition that the point is ( \frac{54}{\sqrt{10}} ) units away from ( M ) to find the possible coordinates of the third corner.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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