Question

1. let the left point be point a and the right point be point b. if you wanted to travel 3/5 of the way from point a to point b, what are the coordinates of the location you travel to? (3 pt)
2. using the same graph in #11, if you travel from point a to point b, and then go from point b up to a new point c, located at (1,5), and then move back to point a, what is the total distance traveled? (3 pt)
3. locate a point that is √2 units away from (-1,5). (3 pt)
4. let point a be located at (6,16) and point b be unknown. if you travel 4/7 of the way to point b, what is the potential location of point b? (3 pt)

Explanation:

1. Question 11:

• First, we need to find the coordinates of points A and B from the graph. Let's assume point A is \((x_1,y_1)\) and point B is \((x_2,y_2)\). From the graph, assume \(A=(- 4,4)\) and \(B=(2,-2)\).
• The formula to find the point \(P=(x,y)\) that is \(t = \frac{3}{5}\) of the way from \(A=(x_1,y_1)\) to \(B=(x_2,y_2)\) is given by:
• \(x=x_1+t(x_2 - x_1)\) and \(y=y_1+t(y_2 - y_1)\)
• \(x=-4+\frac{3}{5}(2+4)\)
• First, calculate \(2 + 4=6\), then \(\frac{3}{5}\times6=\frac{18}{5}=3.6\), and \(-4 + 3.6=-0.4\)
• \(y=4+\frac{3}{5}(-2 - 4)\)
• First, calculate \(-2-4=-6\), then \(\frac{3}{5}\times(-6)=-\frac{18}{5}=-3.6\), and \(4-3.6 = 0.4\)
• So the coordinates of the point are \((-0.4,0.4)\)

1. Question 12:

• First, find the distance between \(A=(-4,4)\) and \(B=(2,-2)\) using the distance - formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
• \(d_{AB}=\sqrt{(2 + 4)^2+(-2 - 4)^2}=\sqrt{6^2+(-6)^2}=\sqrt{36 + 36}=\sqrt{72}=6\sqrt{2}\)
• Then find the distance between \(B=(2,-2)\) and \(C=(1,5)\)
• \(d_{BC}=\sqrt{(1 - 2)^2+(5 + 2)^2}=\sqrt{(-1)^2+7^2}=\sqrt{1 + 49}=\sqrt{50}=5\sqrt{2}\)
• Then find the distance between \(C=(1,5)\) and \(A=(-4,4)\)
• \(d_{CA}=\sqrt{(-4 - 1)^2+(4 - 5)^2}=\sqrt{(-5)^2+(-1)^2}=\sqrt{25 + 1}=\sqrt{26}\)
• The total distance \(D=d_{AB}+d_{BC}+d_{CA}=6\sqrt{2}+5\sqrt{2}+\sqrt{26}=11\sqrt{2}+\sqrt{26}\approx11\times1.414 + 5.099=15.554+5.099 = 20.653\)

1. Question 13:

• Let the point \((x,y)\) be \(\sqrt{2}\) units away from \((-1,5)\). Using the distance formula \(d=\sqrt{(x + 1)^2+(y - 5)^2}=\sqrt{2}\)
• Squaring both sides, we get \((x + 1)^2+(y - 5)^2=2\)
• One possible solution: Let \(x+1 = 1\) and \(y - 5=1\), then \(x=0\) and \(y = 6\). So one point is \((0,6)\)

1. Question 14:

• Let point \(A=(x_1,y_1)=(6,16)\) and point \(B=(x_2,y_2)\)
• Let the point \(P\) that is \(t=\frac{4}{7}\) of the way from \(A\) to \(B\) have coordinates \((x,y)\). We know \(x=x_1+t(x_2 - x_1)\) and \(y=y_1+t(y_2 - y_1)\)
• Let's assume the point \(P\) has coordinates \((x,y)\). We don't know \(P\) exactly, but we can work backward. Let \(x\) and \(y\) be the coordinates of the point after traveling \(\frac{4}{7}\) of the way.
• If \(x=x_1+t(x_2 - x_1)\), then \(x - x_1=t(x_2 - x_1)\), and \(x_2=x_1+\frac{x - x_1}{t}\)
• Let's assume the point after traveling \(\frac{4}{7}\) of the way is some known - point (not given in the problem, so we'll use the general form). But if we assume the end - point of the \(\frac{4}{7}\) journey is the origin \((0,0)\) (for the sake of finding a possible \(B\))
• \(x_2=6+\frac{0 - 6}{\frac{4}{7}}=6-\frac{42}{4}=6 - 10.5=-4.5\)
• \(y_2=16+\frac{0 - 16}{\frac{4}{7}}=16-\frac{112}{4}=16 - 28=-12\)

Step1: Identify coordinates of A and B for Q11

Assume \(A = (-4,4)\), \(B=(2,-2)\)

Step2: Use section - formula for Q11

\(x=-4+\frac{3}{5}(2 + 4)\), \(y=4+\frac{3}{5}(-2 - 4)\)

Step3: Calculate distance between points for Q12

Use \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) for \(AB\), \(BC\), \(CA\)

Step4: Apply distance formula for Q13

\((x + 1)^2+(y - 5)^2=2\), find a solution

Step5: Use section - formula in reverse for Q14

\(x_2=x_1+\frac{x - x_1}{t}\), \(y_2=y_1+\frac{y - y_1}{t}\)

Answer:

1. \((-0.4,0.4)\)
2. \(11\sqrt{2}+\sqrt{26}\)
3. \((0,6)\) (one possible point)
4. One possible location: \((-4.5,-12)\)