Linear functions and geometry problems

Denote by $\tilde{f}$ and $\tilde{g}$ the reflected graphs.

1. From the figure we can see that the original slopes change the sign, and the new lines have the same $y$-intercepts as the old ones. Thus, $\tilde{f}(x)=-3x, \tilde{g}(x)=\frac{1}{2}x-1$
2. From the figure we can see that the original slopes change the sign, and the same is true for the $y$-intercepts. Thus, $\tilde{f}(x)=-3x, \tilde{g}(x)=\frac{1}{2}x+1$
3. From the figure we can see that the original slopes do not change the sign, but the $y$-intercepts change the sign, thus $\tilde{f}(x)=3x, \tilde{g}(x)=-\frac{1}{2}x+1$

Another approach could be to reflect two points of the original graph $f$, and then find the function equation of the new straight line $\tilde{f}$ that passes through these two reflected points.

Line $g$ has the same slope as $f$, $a=-2.5$, thus $g(x)=-2.5x+b$.

1. $y$-intercept $b=0$, thus $g(x)=\underline{-2.5x}$
2. Because $P(4\vert 1)$ is on the graph of $g$ it follows $g(4)=1$. Thus $-2.5\cdot 4+b =1$ and therefore $b=11$. It follows $g(x)=\underline{-2.5x+11}$.
3. First, let's find the $x$-intercept of $f$: Find $x$ with $f(x)=0\rightarrow -2.5x+4=0\rightarrow x=1.6$. The graph of $g$ crosses the $x$-axis one unit to the right of $x=1.6$, so at $x=2.6$. Thus, $g(x)=-2.5x+b$ with $g(2.6)=0 \rightarrow -2.5\cdot 2.6+b=0$, and it follows $b=6.5$ and $g(x)=\underline{-2.5x+6.5}$

Denote the line passing through $A$ and $B$ by $h$. From the figure (see below) it follows that $g$ and $h$ must form a right angle, otherwise $B$ will not be the point closest to $A$. So let us calculate the coordinates of point $B$ by first finding the function equation of $h$, and then intersection $h$ and $g$.

Function equation of $h$: As $h$ and $g$ form a right angle, the slope of $h$ must be (see figure below, left)

We thus have

As $h$ passes through $A(3\vert 5)$, we also have

Coordinates of $B$: We intersect $h$ and $g$, thus we have to find an $x$ with

and therefore the $y$-coordinate is

Thus we have $B(5\vert 0)$. You might have guessed these coordinates from the drawing, but you should be able to calculate them as well!

Distance between $A$ and $B$: Now we can use the theorem of Pythagoras to calculate the distance from $A$ to $B$ (see figure below, right): $d=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{5^2+2^2}=\underline{\sqrt{29}}$.

Denote the line passing through $B$ and $C$ by $f$ (see figure below).

The two lines $f$ and $h$ intersect at point $P$, and form a right angle. So let us first find the function equation of $f$, and then the function equation of $h$.

Function equation of $f$: We have $f(x)=ax+b$, with

The slope of line $f$ is $\frac{9}{12}=0.75$, because

and because $B$ is on the straight line, we have

Thus we have

Function equation of $h$: We have $h(x)=ax+b$ and because $f$ and $h$ are orthogonal, we have (see figure)

Thus it is

And because $A$ is on $h$, we have

and thus it is

Area of triangle: We need the height of the triangle, and the length of the base. The height can be found by determine the distance from $A$ to $P$, where $P$ is the intersection point between $f$ and $h$. So find $x$ with $f(x)=g(x)$, thus $0.75x = -\frac{4}{3}x$, and it follows $x=0$, and therefore $y=f(0)=0$. Thus $P$ has the coordinates $P(0\vert 0)$.

The distance between $A$ and $P$ can be found using the theorem of Pythagoras:

The length of the base $b$ can also be found using the theorem of Pythagoras:

Thus, the triangle area is

Assume one walker starts at the origin ($0 km$), the other at $3.2 km$ from the origin. As we know the walking speed, we can determine where each walker is at every point in time $t$: walker 1 has position $f(t)=4.5\cdot t$, and walker 2 has position $g(t)=3.2-3.5t=-3.5t+3.2$ (see figure below). They meet at the intersection point of the two graphs, because there both walkers have zero distance between them.

Thus to out when that is, we have to find a value for $t$ with

Where do they meet? At position

$s=f(0.4)=4.5\cdot 0.4=\underline{1.8km}$ from the origin.

Draw the situation!!!

1. First, find the slopes of the triangle sides $AB$ and $BC$: $$\text{slope }a_{AB} = \frac{4}{11}$$ $$\text{slope }a_{BC} = \frac{-11}{4}$$ We see that $$a_{BC} = -\frac{1}{a_{AB}}$$ and therefore the lines form a right angle. Thus, the triangle has a right angle at $B$.
2. As the triangle has a right angle at $B$, the area is $$Area=\frac{\vert AB \vert \cdot \vert BC \vert}{2}$$ where $\vert AB \vert$ denotes the length of triangle side from $A$ to $B$, and $\vert BC \vert$ is the length of the triangle side form $B$ to $C$. We can use Pythagoras to find those lengths: $$\vert AB \vert = \sqrt{11^2+4^2}=\sqrt{137}$$ $$\vert BC \vert = \sqrt{(-11)^2+4^2}=\sqrt{137}$$ Thus, the area is $$Area=\frac{\sqrt{137}\cdot \sqrt{137}}{2}=\frac{137}{2}=\underline {68.5}$$
3. We know how to get from point $B$ to point $C$, just move along the slope triangle between $B$ and $C$, e.g.: $$\Delta x= -4, \Delta y=11$$ In the same way we can get from $A$ to $D$. So starting at $A$, move $4$ to the left and $11$ upwards. We get the point $$D(1-4\vert -3+8)=\underline{D(-3\vert 8)}$$
4. Denote the diagonal from $A$ to $C$ by $f$, and the diagonal from $B$ to $D$ by $g$. Let's find the function equation of the diagonals, so that we can intersect them. $$f(x)=ax+b$$ $$\text{slope } a = \frac{\Delta y}{\Delta x} = \frac{15}{7}$$ Thus, $$f(x)=\frac{15}{7} x+b$$ To find $b$, we use that $A$ is on the graph of $f$: $$f(1)=-3\rightarrow \frac{15}{7} \cdot 1+b=-3 \rightarrow b=-\frac{36}{7}$$ Thus, we get $$f(x)=\underline{\frac{15}{7} x-\frac{36}{7}}$$ Similar, to find the function equation of $g$, we set $$g(x)=ax+b$$ $$\text{slope } a = \frac{\Delta y}{\Delta x} = \frac{-7}{15}$$ Thus, $$g(x)=-\frac{7}{15} x+b$$ To find $b$, we use that $B$ is on the graph of $g$: $$g(12)=1\rightarrow -\frac{7}{15} \cdot 12+b=1 \rightarrow b=\frac{99}{15}$$ Thus, we get $$g(x)=\underline{-\frac{7}{15} x+\frac{99}{15}}$$ Finally, we have to solve the equation $$\frac{15}{7} x-\frac{36}{7} = -\frac{7}{15} x+\frac{99}{15}$$ to get the $x$-coordinate of the intersection point between $f$ and $g$. $$\begin{array}{llll} \frac{15}{7} x-\frac{36}{7} &=& -\frac{7}{15} x+\frac{99}{15} & \vert \cdot 7\\ 15 x-36 &=& -\frac{49}{15} x+\frac{693}{15} & \vert \cdot 15\\ 225 x-540 &=& -49 x+693 & \vert +540, +49x\\ 274 x &=& 1233 & \vert :274\\ x &=& 4.5 \end{array}$$ The $y$-coordinate is $$y = f(4.5)=\frac{15}{7} \cdot 4.5-\frac{36}{7} = 4.5$$ Thus, we have $\underline{\underline{S(4.5\vert 4.5)}}$.

Thus $f(x)=2.21739...\cdot x +b$. Because of $f(2)=0$ follows

Thus, $f(x)=2.21739...\cdot x -4.4347...$.

No computations are shown, just the results.

1. See below
2. Because the have the same slope.
3. See below
4. See below
5. See below
6. See below
7. The fraction is always $2$.