Functions Questions…

Question

Hey im doing some functions questions and im looking at my notes and i cant figure out how to do them. I also have answered the first 3 and i would appreciate if anyone could tell me if i am correct...

1.  Let $p_1$ = (-12, 10) and $p2$ = (14,-10). Find:
i.  the Euclidean distance between p¬1 and p2 = 32.80
ii. the Manhattan distance between p¬1 and p2 = 46
iii.    the mid-point between p¬1 and p2 (1,0)

2.  Let $p1$ = (5, -7, 4) and $p2$ = (3, 6, -4). Find:
i.  the Euclidean distance between p¬1 and p2 = 69
ii. the Manhattan distance between p¬1 and p2 = 23
iii.    the mid-point between p¬1 and p2 = 

3.  Which of the following are equations of lines:
i.  2 x + 5 y = -2  True
ii. –5.3 – 9.7 x2 + 4 y = -3 False
iii.    y = 0.75 x – 1.4  True

4.  What is the slope of each of the lines in question 4?

5.  The slope of a line that passes through the point (2,3) is 4. What is the equation of the line?

6.  Find an equation of the line that is perpendicular to the line -2 x - 6 y = -18.4 and passes 
through the point (4,-6).

 7.  Determine the equation of the curve of shortest distance which joins the point (2,-1) to the 
      line 2x-5y= - 4. 

Answer 1

1. 1. Euclidian distance is $\sqrt {(14 - (-12))^2 + (-10-10)^2} = \sqrt{26^2+20^2}\approx32.80$, so you're right.
   2. Manhattan distance is $\left|14-(-12)\right|+\left|-10-10\right| = 46$, so you're right there, too.
   3. Midpoint is $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) = (\frac{14-12} {2},\frac{-10+10}{2})=(1,0)$, so that's also correct.
2. 1. $\sqrt{(5-3)^2+(-7-6)^2+(4-(-4))^2}=\sqrt{237}\approx15.395$
   2. 23 is right.
   3. The midpoint is $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}) =(\frac{5+3}{2},\frac{-7+6}{2},\frac{4-4}{2})=(4,\frac12,0) $
3. Your answers are correct.
4. Your question makes no sense; no lines are given in number 4.
5. Since you have a point and the slope of the line, you can use point-slope form: $$ \begin{align} y-y_0 &= m(x-x_0)\\ y-3&=4(x-2)\\ y&=4x-5 \end{align} $$
6. The slope of the given line is $-\frac13$, so the slope of the perpendicular line is the opposite reciprocal. That is, $m=3$. So, using the point-slope form again, $$ y-(-6)=3(x-4)\\ y=3x-18 $$
7. If you're just asking for the shortest distance to the line, you can use the formula $$ \frac{\left|Ax_1+By_1+C\right|}{\sqrt{A^2+B^2}}, $$ which gives $\frac{13}{\sqrt{29}}\approx2.41$