3.1 The Geometry of Lines

A line of 1 dimension in the xy-plane can be expressed as y=mx+b, which is the slope (m) - intercept (b) form.  There is also the point-slope form: y-(Subscript[y, 0]) = m(x-(Subscript[x, 0])), where m is once again the slope, and ((Subscript[x, 0]), (Subscript[y, 0])) is a point on the line, i.e. it satisfies the equation (Subscript[y, 0]) = (Subscript[mx, 0])+b.  Lastly, there is standard form: Ax+By=C, where both A and B cannot be zero.

L0[x_]:=2x-3

(L0[12]-L0[4])/(12-4)

(L0[2]-L0[5])/(2-5)

L1[x_]:=-0.3x+2.7

(L1[6.2]-L1[2.3])/(6.2-2.3)

(L1[15.2]-L1[-3.8])/(15.2-(-3.8))

L[x_]:=m*x+b

L[x1]-L[x0]

%/(x1-x0)

Simplify[%]

Plot[{3*x+2,-3x+2,3*x-2,-3*x-2},{x,-5,5}]

Example 3.1

The original information was that for a 1 ft horizontal run, the vertical rise of the roof was 0.5 feet...

Part 1

In Figure 3.5, there is a 14 ft run from the edge of the house to a line directly under the peak of the roof, we know the roof rises 0.5 ft for each horizontal foot.  Thus, the height of the roof gains 7 feet over this run.  If the roof starts at 8ft off the ground, then we are at 8+7 = 15 feet off the ground.

Part 2

The question is worded strangely, but if we continue the roof down to the ground off the left side of Figure 3.5, then we should be able to determine the distance on the line from the end of the roof to the ground makes from the base of the house.  Since each linear foot to the left gives a 0.5 drop downwards, we will have to go 16 steps of 0.5 feet to get to the ground vertically, thus 16 feet is the horizontal distance.

Example 3.2

Part 1

The slope of the line that follows the circus tent is simply m = (rise/run)= (2/5) since the height above ground goes from 10 to 12 ft during a 5 ft horizontal run.

Part 2

To find the height of the tent at its center pole, we need to multiply the slope by 60, then add 10, which gives 34 ft:

2/5*60+10

Part 3

Attaching a rope and following the slope of the tent roof to the ground, we know that the (rise/run)  must be the same thus:
(2/5) = (10/x)
We can cross multiply, or use the Solve command, the result is 25 ft.

Solve[2/5==10/x,x]

Exercise 10

To determine how tall the statue is, we simply need to determine a slope.  Note the one slope we have is a rise of 35 feet to a run of 600 feet.  So m = (rise/run) = (35/600).  The slope of the line passing from the statue to the building base must also be the same so we set up the equation:
(35/600)= (h/125)
Using the solve command below, we see that the statue is approximately 7.3 ft tall. Ask yourself, does this make sense?

Solve[35/600==h/125,h]

N[%]

Exercise 20

Part a.

If we are starting from Lake Itasca and looking down towards the Gulf of Mexico, then rise over run gives
m = -(1475/2340)(ft/mi), if we are looking from the Gulf of Mexico to Lake Itasca, we have m =(1475/2340)(ft/mi).  This can be seen in the top figure in the sketches above.

Part b.

See the middle pic in the above sketch. Memphis is 1982 miles downriver from Lake Itasca, which means it is 358 miles upriver from the Gulf.  We can use the 358 miles from the Gulf to compare slopes:
(1475/2340) = (y/358) , where y is the height above sea level.  We can solve this equation using the Solve command to get that Memphis is 226 feet above sea level.

Solve[1475/2340==y/358.0,y]

Part c.

For this last part, we need to figure out how far from Lake Itasca a spot 200 ft is above sea level.  See the bottom pic above.  We can use the same slope idea again, in this case, we have  (1475/2340) = (200/x).

Clear[x]

Solve[1475.0/2340.0==200.0/x,x]

1475.0/2340

200/317.288

So the spot 200 ft above sea level is 317 miles from the Gulf, but we need the distance to Lake Itasca, which is 2340-317 = 2023 miles.

2340-317