# Slope

This week’s topic is:  Distance, Midpoint, and Slope.

### Q:  The slope of the line passing through (-1,-3) and (7,y) is -1/2.  What is the value of y?

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Algebraic solution by hand:

Use the slope formula:

Solution using my TI-84 Plus program DISTANCE:

Run DISTANCEHow?

Use your multiple choice answer choices.  One of them is -7.  (If you don’t have choices, draw a picture.  Plot the point (-1,-3) and do a slope of -1/2 from there – down 1, right 2, down 1, right 2, … until you get to a point where x is 7.  Read the y-value of that point – it will be -7 and check it using this program.)

x1=-1

y1=-3

x2=7

y2=-7

Ignore the distance information – we don’t need it for this problem.  Press ENTER for the midpoint information, which we don’t need either.  Press ENTER once more to find that the slope is -1/2.

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# Distance

This week’s topic is:  Distance, Midpoint, and Slope.

### Q:  For what value(s) of x is the point (x,4) exactly 5 units away from the point (6,8)?

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Solution by hand:

Use the distance formula:

Solution using my TI-84 Plus program DISTANCE:

Draw a picture:

Guess that x=2, maybe.  It’s probably an integer if this is an SAT problem.  Use the answer choices if available.

Run DISTANCEHow?

x1=2

y1=4

x2=6

y2=8

The distance is 5.66.  It’s too big – you wanted it to be 5 – so move the point a little closer to the middle.

Maybe x=3:

x1=3

y1=4

x2=6

y2=8

The distance is exactly 5.  Perfect.  (There’s also another possible answer, x=9.  If you have multiple choice answers, you will see 3 and 9 as a choice, so try both and see that they both work.  For a grid-in SAT problem, you would only have to find one answer anyway.)           :

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# SAT Question Of The Day – estimating for hard geometry

### You’d be surprised how often you can really do this and get the right answer.  Follow me, and you’ll see lots of these!

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# ACT, Geometry – diagonal of a rectangular prism

Q: What is the (straight-line) distance between points A and B on the figure below?

Explanation:  We are finding the length of the green segment below:

This requires the three-dimensional version of the distance formula:

There is another way to do this, using a progression of two right triangles, with the green segment being the hypotenuse of the second triangle, but I find the above formula much simpler.  It is just an extension of the ‘usual’ (2-dimensional) distance formula we are so familiar with from Algebra and Geometry.

BUT if you want an even easier way, and I know you do, use my TI-84 Plus program ‘DIST3D’.

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# Geometry – polygons

Q: A hexagon has interior angles measuring (in degrees) 137, 100, 150, 114, 90, and x. What is the value of x?

Explanation:
The total measure of all the interior angles in a polygon is (n-2) * 180, where n is the number of sides of the polygon.
A hexagon has 6 sides, so n = 6. The total is (6-2) * 180 = 720.
Add up the five given angle measures: 137 + 100 + 150 + 114 + 90 = 591.
The measure of the remaining angle is 720 – 591 = 129.
My TI-84 Plus program POLYGON will find this total for you.
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# Geometry – volume

Q: If 20 solid gumballs with diameter 1 inch are placed inside a hollow sphere of diameter 6 inches, how much empty space is there in the larger sphere?

Explanation:

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# SAT, Geometry – triangles

Q: The lengths of two sides of a triangle are 4 and 6. If the third side is an integer, what is one possible value for the perimeter?

Answer:  The perimeter can be any of the following values: 13, 14, 15, 16, 17, 18, or 19.

Explanation: The length of the third side of any triangle must be longer than the difference of the other two sides, but shorter than the sum. (This is called the Triangle Inequality Theorem.) Here, the third side is longer than 6-4=2 and shorter than 6+4=10. It can be any integer between, but not including, 2 and 10. Recall that integers include only whole numbers and their negatives. The third side must then be either 3, 4, 5, 6, 7, 8, or 9. To get the possible perimeters, add the other two side lengths (4 and 6) to each of the possibilities for the third side. The result is that the perimeter can be 13, 14, 15, 16, 17, 18, or 19.

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# Geometry – Areas

Q: What is the area of this figure?

Explanation:  Cut the figure into two rectangles as shown.  Find the missing vertical length by noticing that the two shorter vertical lengths must add up to the long vertical length.  The missing vertical length is 3, because 4 + 3 = 7.  Do the same for the missing horizontal length.  It is 5, because 6 + 5 = 11.

Now, find the area of each rectangle (A = length x width).  The one on the left is 7×6 = 42.  The one  on the right is 3×5 = 15.  Add these areas to find the area of the original figure:  42 + 15 = 57.

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# Geometry – Lines and segments

Q: Points D, E, and F are collinear with DE < EF and DF < EF.  If DE = 2x+6, DF = 3x-4, and EF = 22, what is DE?

Explanation:

First, recall that ‘DE’ refers to the length of the segment with endpoints D and E.  It is the distance between D and E.  Next, DON’T ASSUME that E is between the D and F!  Since DE and DF are each less than EF, and all three points lie on the same line (collinear), D must be between E and F.

Draw the diagram:

__________________________

E                         D                   F

Now label the distances as they are given in the problem:

<————–22——————->

__________________________

E        2x+6         D     3x-4      F

Use the concept: Part + Part = Whole (also called the Segment Addition Postulate) to write the equation:

2x+6 + 3x-4 = 22

Solve:

5x + 2 = 22

5x = 20

x = 4

Plug x=4 back in to the ‘DE’ part:

DE = 2x+6 = 2(4) + 6 = 14.

Tip:  To be sure your answer is correct, plug x=4 into the ‘DF’ part, too:

DF = 3x-4 = 3(4)-4 = 8.  Now, see that the sum of the two ‘parts’ does equal the ‘whole’:

14 + 8 = 22.