Author: drisebor@hotmail.com

The corners of a triangle are plotted on a Cartesian plane below

1. Find the gradient of each side
2. Find the x and y intercepts of each side
3. Which is the longest side?
4. Find the mid point of the longest side
5. Move one of the corners up, down, left or right by one place (keeping the x or y value as an integer) to turn the triangle into an Isosceles triangle. How many ways can the Isosceles triangle be created?

 

Here’s a super ugly, messy looking expression to simplify. Good luck!

Tip 1: Don’t just try to expand the whole thing out.

Tip 2: Tidy up the numerator by separating into multiples of either of the two factors in the denominator. Then the expression can be converted into two fractions.

Tip 3: Identify the perfect square and difference of squares.

 

 

In my previous post I described a shape mapped to a Cartesian plane. The shape was described using a series of connected points.

Question 1

The pairs of connected points made lines which are shown in the diagram below with the answer to question 1, the gradient of each line, shown as the value of M (from y=Mx + C)

The gradient can be found from each pair of points, since the gradient is the difference in y divided by the difference in x for two points, ie

We also note that there is a lot of symmetry going on in the diagram. The shape is symmetric around both the x and y axis, so we really only need to calculate two gradients with this equation. The rest can be calculated by inverting  the gradient for lines on the opposite side of the x or y axis, while the gradient is the same for lines that are on the opposite side of the shape.

Question 2

The y intercept of 4 of the 8 lines is already given in one of the point coordinates of lines (h),(a),(e) and (d).

For the other 4 lines, the y intercept may be calculated using:

y=Mx + C,

so y = C when x=0,

C is a constant so we just need to find the value of C, which is easy if we have the x intercept coordinate:

C = y -Mx, y=0 (because it’s the x intercept)

so C = -Mx, where x is the x coordinate of the x intercept

In practical terms this means that the magnitude of the y intercept is M times the magnitude of the x intercept so that for line (b), for example, the y intercept is y = 3 x 4 = 12,

so the equation for (b) is y = -3x + 12

Once we have the y intercept for one of the lines, the rest are easy due to symmetry, so

Equation	y-intercept
(a)	4
(b)	12
(c)	-12
(d)	-4
(e)	-4
(f)	-12
(g)	12
(h)	4

Question 3

The shape is symmetrical along the x and y axis, and further examination shows that it is also symmetrical along y=x and y=-x. This means that all lines are the same length.

We use Pythagoras, so

Question 4

Opposite lines are parallel (M is the same), lines with M=3 and M=-1/3 are perpendicular and similarly lines with M=-3 and M=1/3 are also perpendicular. Seeing the pattern??

Question 5 and 6

The diagram below is a bit of a dog’s breakfast, but nevertheless shows the midpoints and the perpendicular lines on the intercepts.

To find the equations, the next step is to find the y intercepts, since finding the gradients should be trivial by now.

We can find the y intercept using C = y -Mx

eg for equation (a),

Therefore equation (a) is y = 3x -1

The equations are presented in the following table:

	equation
(a)	y=3x-1
(b)	y=1/3 x + 1/3
(c)	y=-1/3 x – 1/3
(d)	y=-3x +1
(e)	y = 3x +1
(f)	y = 1/3x – 1/3
(g)	y = -1/3x  + 1/3
(h)	y = -3x -1

Question 7

The shape is an octagon.

Stretch Question

The octagon would be regular if all sides and angles were the same, however this can’t be the case because while the points that are on one of the axes are all 4 away from the origin, the other points (eg (3,3) ) are 3 sqrt(2) away from from the origin.

One way to make the octagon regular would be to move the points that are along the axis so that the magnitude of the non zero (axis intercept) coordinate is 3 sqrt(2). Another way would be to move the points that have x=y so that they are exactly 4 from the origin, while maintaining x=y.

To do this, use Pythagoras’ rule, ie:

so adjust the magnitude of x and y to be 2 sqrt(2)

If you did this, all the pairs of perpendicular lines in the previous diagram should merge and intersect the origin. Want to try it??

The following lines on a cartesian plane make the outline of a shape:

(0,4) to (3,3)
(3,3) to (4,0)
(4.0) to (3,-3)
(3,-3) to (0,-4)
(0,-4) to (-3,-3)
(-3,-3) to (-4,0)
(-4,0) to (-3,3)
(-3,3) to (0,4)

1) Find the gradient of each line.

2) Find the y intercept if not known

3) Find the length of each line. (Are all the lengths the same? Shortcuts are allowed.)

4) Which lines are parallel and which are perpendicular

5) For each line, find the midpoint and find the equation for the line that is perpendicular.

6) For each of the perpendicular lines found, identify the y intercept.

7) What type of shape is outlined by the original lines?

Stretch question:

Is this the ‘regular’ version of the shape? If not, what would be required to convert it to the regular version.

Try expanding these binomials, (Don’t use an air pump or they will explode and turn into power series – jk 😉 Instead use FOIL – First, Outside, Inside, Last)

There is something interesting about the last four expansions. What do they have in common and how can you spot it without expanding the expression?

Here are some more fractions I came up with to practice on.

Here are my answers. Let me know if you think there are any mistakes!

Here’s an trigonometry elevation problem to keep you occupied:

Bob the rock climber wants to know how far left until he gets to the summit of the sheer (vertical) rock face. He is communicating with his friend Jerry on the ground who has worked out that from his (Jerry’s) point of view the top of the cliff is 100 meters away. Jerry also finds out that the angle of elevation of the top of the cliff is 55 degrees and the angle of elevation of Bob is 25 degrees. How far does Bob have left to travel?