How many pairs of vertical angles are formed by five distinct lines that have a common point of intersection?

Mathematics

1 answer:

Bess [88]4 years ago

7 0

25 pairs of vertical angles are formed.

<u>Step-by-step explanation:</u>

Suppose there are N distinct lines, which produces N squared pairs of vertical angles.

If we have 2 distinct lines, then we have 4 pairs of vertical angles are formed.

2 distinct lines, then 2² = 4 pairs of vertical angles

3 distinct lines, then 3² = 9 pairs of vertical angles

4 distinct lines, then 4² = 16 pairs of vertical angles

Here we have 5 distinct lines, and these lines produces 5² pairs of vertical angles.

So here 25 pairs of vertical angles are formed by 5 distinct lines.

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The midpoint of AB is at (3,7) and A is at (0,-5). Where is B located?

olga nikolaevna [1]

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
% (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)\qquad thus
\\
----------------------------\\$ $\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 0}}\quad ,&{{ -5}})\quad 
% (c,d)
B&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint 
(3,7)\impliedby midpoint\qquad thus
\\ \quad \\
\left(\cfrac{{{ x_2 }} + {{ 0}}}{2}=3\quad ,\quad \cfrac{{{ y_2 }} + {{( -5)}}}{2}=7 \right)\to 
\begin{cases}
\cfrac{{{ x_2 }} + {{ 0}}}{2}=3
\\ \quad \\
\cfrac{{{ y_2 }} + {{ -5}}}{2}=7
\end{cases}
\\ \quad \\
solve\ for\ x_2\ and\ y_2$