a) 1) G(8,9), 2) G(8,7), 3) G(8,12), 4) G(7,5), 5) G(4,5), 6) G(4,4)
b) None of them have two edges connecting the same two vertices.
c) 1, 2, 3, all have a possible path between every two vertices. 4 doesn't, there are two clear parts in the graph. 5 is directed, and you can find a path between each two vertices. 6 isn't connected, but the underlying undirected path is.
CloseEach person shook hands with 5 people. \(\sum_{\nu \in V} \operatorname{deg}(\nu) = 2e\), and there are \(6 \times 5\) degrees in the graph, so the number of edges is \(e = \frac{30}{2} = 15\).
Closea) 1. \(\left(\begin{array}{llll}0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0\end{array}\right)\)
a) 2. \(\left(\begin{array}{lllll}0 & 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0\end{array}\right)\)
a) 3. \(\left(\begin{array}{lllll}0 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 & 0\end{array}\right)\)
b) 1.
b) 2.
b) 3.
1. in: 2 out: 2
2. in: 1 out: 2
3. in: 2 out: 1
4. in: 3 out: 1
5. in: 0 out: 2
6. in: 3 out: 2
Closea) Yes
b) There are multiple possibilities, such as: ABECA, ABGFBA, ABGFCA, ACEBA...
c) FGBACEBFCD
Closea) Every vertex has a path to every other vertex.
b) The graph has 6 vertexes. We need each vertex to gave a degree \( \geq \frac{6}{2}=3 \). Each vertex has at least a degree of three, so according to Dirac's theorem, we know there is a Hamiltonian path.
c) Not all vertices have an even degree in this graph, so it is not a Eulerian circuit. The ones that don't are: \( D \), \( B \), \( C \), and \( F \).
d) We add edges to the vertices with an uneven degree, so that they all have an even degree. We can add an edge between \( C \) and \( B \), then \( D \) and \( F \). Then, every vertex has an even number of edges.
a) You can place the y-intercept and any other point of your choice to get the following graph:
b) By adding the line \(x = 2\), we get the following:
c) Place the symmetry of two points on \(L_1\), and draw a line through them.
d) You can use the points you used to draw \(L_2\) to find the equation. For example, the symmetry of \((0, 2)\) is \((4, 4)\), and the symmetry of \((4, 0)\) is \((0, 0)\). This gives us the gradient \(\frac{4 - 0}{4 - 0} = 1\) and the intercept \((0, 0)\).
e) \([AB]\) and \([AC]\) are symmetrical, so they have the same length.
a) \(y=2(0)+5=5\)
b) \(0=2x+5\), so \(x=-\frac{5}{2}\)
c) 2
d) The product of the gradients must be -1. Hence it is \(-\frac{1}{2}\).
Closea) First we need to find the slope of \(L_1\). This is \(\frac{\Delta y}{\Delta x} = \frac{8-2}{-5-1}=-1\). Then we can plug into the point-slope form to find the equation of the line.
\[ (y-2)=-1(x-1) \\ y = -x +3 \]
b) First, we can find the midpoint.
\[ M = \left(\frac{1-5}{2},\frac{2+8}{2}\right) = \left(-2,5\right) \]
Then we remember that two lines are perpendicular if and only if the product of their slopes is -1. Since \(L_1\) has a slope of -1, \(L_2\) must have a slope of 1.
\[(y-5) = 1(x+2) \\ y= x+ 7 \]
Closea) \(17=5p+2\), so \(p=3\)
b) The gradient of \(L_2\) must be \(-\frac{1}{3}\) as \(3\cdot (-\frac{1}{3}) = -1\), which needs to be the case for perpendicular lines. Using point-slope form:
\[(y-17) = -\frac{1}{3}(x-5) \\ y = -\frac{1}{3}x + \frac{56}{3} \]
c) The equation in standard form is \(3y + x - 56 = 0\)
a) \(C : \left(\frac{2+4}{2},\frac{6+4}{2}\right) = \left(3,5\right)\)
b) \(-2(6)+5(2)+2=0\)
c) \(AC = \sqrt{(3-2)^2+(5-6)^2} = \sqrt{2}\)
d) First, we can find the gradient of \(L_1\), so we need to rearrange the equation. \(L_1: y=\frac{5}{2}x+1\). Since \(L_2\) will be perpendicular, their gradients' product must give \(-1\), hence the gradient of \(L_2\) is \(-\frac{2}{5}\).
e) We can use point-slope form: \[ (y-5)=-\frac{2}{5}(x-3) \\ y = -\frac{2}{5}x + \frac{31}{5} \]
f) This is a system of two equations: \[ y=\frac{5}{2}x+1 \\ y = -\frac{2}{5}x + \frac{31}{5} \] Solving this for \(x\) and \(y\) gives \(x=\frac{52}{29} \approx 1.79\) and \(y=\frac{159}{29} \approx 5.48\).
