Month: October 2018

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??