Solutions Part B:

11) A line and a parabola intersect at points A and C. Their equations are:

5x – 4y + 48 = 0 and y = 0.25x² – 7x + 41

Line segment AC is the hypotenuse of right triangle ABC, with legs // to the axes.
What are the coordinates of point B?

Solution: We need the x value at A and the y value at C for the coordinates of B.

First, find the points of intersection of the 2 curves: (I use ¼ instead of 0.25)

From 5x – 4y + 48 = 0, we get and set it = ¼ x² – 7x + 41

Multiply through by 4 and collect terms to get: x² – 33x + 116 = 0

This factors as: (x – 4)(x – 29) = 0.

The x-value at A = 4

To find the y value at C, set x = 29 in either equation: (line is easier)

The coordinates of B are (4, 48.25)

12) What is the result of (6x³y³ – 11x²y² + 18xy – 5) ÷ (3xy – 1)

Soln:

13) In the Cartesian plane, function f is a straight line that passes through A(0, 9) and B( t, – 90).
This line is parallel to another line whose rule is g(x) = 3x – 16

What is the value of t?

Soln: since the lines are parallel, their slopes must be equal.

Therefore so t = – 33.

14) The rule of function g is g(x) = p x² + r x – 36, where p and r are not equal to zero.

Function g is positive over the intervals .

What is the value of p?

Soln: We have a parabola opening upwards with zeros at – 6 and 10,

so we substitute – 6 and 10 for x and 0 for g(x) and get 2 equations in p and r :

0 = p ( – 6)² + r ( – 6) – 36 , and 0 = p ( 10)² + r (10) – 36

We turn these into: 36 = 36p – 6r which becomes 6 = 6p – r

and 36 = 100p + 10r which becomes 3.6 = 10p + r

Now we add them together to eliminate "r":

We get 9.6 = 16 p so p = 9.6 ÷ 16 = 0.6

15) Triangle PQR has these characteristics:

angle PQR = angle QRP .............. angle RPQ = 36° .............QR = 47 cm

What is the length of segment PR to the nearest centimetre?

Soln: We have an isosceles triangle with equal sides PQ and PR.

We know the vertex angle is 36° so the 2 other angles are each = ½(180° – 36°) = 72°.

When we draw the perpendicular PS from P to QR, we know that S is the mid-point.
Since QR = 47 cm (given), SR = 23.5 cm.

In triangle PSR, we know that = 76 cm.

16) In quadrilateral ABCD, line segments TU and VW intersect at S.

Points T, U, V and W are on quadrilateral ABCD.
Quadrilaterals AVST and CWSU are congruent.

Complete steps 2 and 3 of the proof that ABCD is a parallelogram.

Step 1	Quadrilateral AVST Quadrilateral CWSU	Given
Step 2	.	Because corresponding angles in congruent figures are congruent.
Step 3	AB // DC AD // BC	Because alternate angles are equal, so lines are parallel.
Step 4	ABCD is a parallelogram	Because the opposite sides of a parallelogram are parallel.

( MathTub Index )