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KIM [24]
3 years ago
11

Two vertices of an equilateral triangle are located at (2, 0) and (–2, –3). What is the triangle’s perimeter?​

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
3 0

Answer:

15

Step-by-step explanation:

one side is equal to 5. So 5 ×3

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6 0
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Number 1d please help me analytical geometry
lesantik [10]
For a) is just the distance formula

\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;A&({{ x}}\quad ,&{{ 1}})\quad &#10;%  (c,d)&#10;B&({{ -4}}\quad ,&{{ 1}})&#10;\end{array}\qquad &#10;%  distance value&#10;\begin{array}{llll}&#10;&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}&#10;\\\\\\&#10;\sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2}&#10;\end{array}
-----------------------------------------------------------------------------------------
for b)  is also the distance formula, just different coordinates and distance

\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;A&({{ -7}}\quad ,&{{ y}})\quad &#10;%  (c,d)&#10;B&({{ -3}}\quad ,&{{ 4}})&#10;\end{array}\ \ &#10;\begin{array}{llll}&#10;&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}&#10;\\\\\\&#10;4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2}&#10;\end{array}
--------------------------------------------------------------------------
for c)  well... we know AB = BC.... we do have the coordinates for A and B
so... find the distance for AB, that is \bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;A&({{ -3}}\quad ,&{{ 0}})\quad &#10;%  (c,d)&#10;B&({{ 5}}\quad ,&{{ -2}})&#10;\end{array}\qquad &#10;%  distance value&#10;\begin{array}{llll}&#10;&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\&#10;d=\boxed{?}&#10;&#10;\end{array}

now.. whatever that is, is  = BC, so  the distance for BC is

\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;B&({{ 5}}\quad ,&{{ -2}})\quad &#10;%  (c,d)&#10;C&({{ -13}}\quad ,&{{ y}})&#10;\end{array}\qquad &#10;%  distance value&#10;\begin{array}{llll}&#10;&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\&#10;d=BC\\\\&#10;BC=\boxed{?}&#10;&#10;\end{array}

so... whatever distance you get for AB, set it equals to BC, BC will be in "y-terms" since the C point has a variable in its ordered points

so.. .solve AB = BC for "y"
------------------------------------------------------------------------------------

now d)   we know M and N are equidistant to P, that simply means that P is the midpoint of the segment MN

so use the midpoint formula

\bf \textit{middle point of 2 points }\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;M&({{-2}}\quad ,&{{ 1}})\quad &#10;%  (c,d)&#10;N&({{ x}}\quad ,&{{ 1}})&#10;\end{array}\qquad&#10;%   coordinates of midpoint &#10;\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P&#10;\\\\\\&#10;

\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies &#10;\begin{cases}&#10;\cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\&#10;\cfrac{{{ y_2}} + {{ y_1}}}{2}=4&#10;\end{cases}

now, for d), you can also just use the distance formula, find the distance for MP, then since MP = PN, find the distance for PN in x-terms and then set it to equal to MP and solve for "x"


7 0
3 years ago
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