one endpoint of AB has coordinates (-3,5) If the coordinates of the midpoint of AB are (2,-6),what is the length of AB?
1 answer:
We know that (-3,5) is the location of one of the endpoints.... and we know the midpoint is at (2,-6)... .now.. what's the distance between those two guys?
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -3}}\quad ,&{{ 5}})\quad % (c,d) &({{ 2}}\quad ,&{{ -6}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ d=\sqrt{[2-(-3)]^2+[-6-5]^2}\implies d=\sqrt{(2+3)^2+(-6-5)^2} \\\\\\ d=\sqrt{5^2+(-11)^2}\implies d=\sqrt{25+121}\implies d=\sqrt{146}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%28%7B%7B%20-3%7D%7D%5Cquad%20%2C%26%7B%7B%205%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%202%7D%7D%5Cquad%20%2C%26%7B%7B%20-6%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B2-%28-3%29%5D%5E2%2B%5B-6-5%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%282%2B3%29%5E2%2B%28-6-5%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B5%5E2%2B%28-11%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B25%2B121%7D%5Cimplies%20d%3D%5Csqrt%7B146%7D)
so, the distance "d" from the midpoint to that endpoint is that much. And the distance from the midpoint to the other endpoint is the same "d" distance, because the midpoint is half-way in between both endpoints.
so, the length of AB is twice that distance, or
so, the distance "d" from the midpoint to that endpoint is that much. And the distance from the midpoint to the other endpoint is the same "d" distance, because the midpoint is half-way in between both endpoints.
so, the length of AB is twice that distance, or
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Answer:
a) the length of the wire for the circle =
b)the length of the wire for the square =
c) the smallest possible area = 126.02 in² into two decimal places
Step-by-step explanation:
If one piece of wire for the square is y; and another piece of wire for circle is (60-y).
Then; we can say; let the side of the square be b
so 4(b)=y
b=
Area of the square which is L² can now be said to be;
On the otherhand; let the radius (r) of the circle be;
2πr = 60-y
Area of the circle which is πr² can now be;
Total Area (A);
A =
=
For the smallest possible area;
∴
If we divide through with (2) and each entity move to the opposite side; we have:
By cross multiplying; we have:
2πy = 480 - 8y
collect like terms
(2π + 8) y = 480
which can be reduced to (π + 4)y = 240 by dividing through with 2
∴ since , we can determine for the length of the circle ;
60-y can now be;
=
=
=
=
also, the length of wire for the square (y) ;
The smallest possible area (A) =
= 126.0223095 in²
≅ 126.02 in² ( to two decimal places)