REVIEW / THE ROAD TO A "NEW" DISTANCE FORMULA

Hey Guys!

Well on Friday's class we had Mr. Clark once again to fill in for Mr. K, and no time was wasted. To kick off the class, Exercise 20: Review 2 was assigned and is expected to be done.

As soon as that was established, our class as a whole had asked Mr. Clark to look over a few questions on Exericse 21: Circles on a Coordinate Plane (which was previously assigned) that we were having problems with.

4. Find the centre and radius, and sketch the graph for the following circles.
(c) x²+y2+6y-12=0
HINT: To solve, complete the square

and put into standard form of the equation of a circle
x²+y²+6y-12=0
x²-10x+25+y²-4y+4=25+4
(x²-5)2+(y-2)²=29
centre: (5,2) radius: √29

9. Sketch the graph of y=x²-2x+5. State the domain and the range.
HINT: The degree is 2, we have a parabola!
y=x²-2x+5
y=(x²-2x+1)+5-1
y=(x-1)²+4

HINT: Vertex is (1,4) and Y-Intercept is (0, 5)
Domain: (-, ∞) Range: [4, ∞)

10. Write a quadractic equation with roots 2±√3
HINT: Find Sum of Roots + Product of Roots
s.o.r. = 2+√3 + 2-√3
=4
p.o.r.= (2+√3)(2-√3)
=4-3
HINT: (√x)(√x)=x
=1
eq'n = x²-4x+1

11. Two sidewalks meet at right angles. At noon, Person A is 12 km north of the intersection, walking south at 2 km/h. Person B is 18 km east of the intersection, walking east at 4 km/h. At what time is the area of AOB a maximum?
HINT: Maximum.. think parabola.. think sign of a quadratic equation.. think vertex
Original Area = Base(Height)/2
= 18 km(12 km)/2
= 108 km²

After X hours A will be 12 km-2x
After X hours B will be 18+4x

New Area = (18+4x)(12-2x)/2
= (216-36x+48x-8x²)/2
= (-8x²+12x+216)/2
= -4x²+6x+108
HINT: lets complete the square to find the vertex
= -4x²+6x+108
= -4[x²-(3/2)x+(9/16)]+108+(9/4)
= -4[x-(3/4)]² + 441/4
Vertex (3/4, 441/4)

So that was all with Exercise 21: Circles on a Coordinate Plane.

Students then asked questions about our next assignment
Exercise 22:Distances between Points and Lines

We were all familar with the line distance formula
distance =√(x₂, x₁)²+(y₂-y₁)²

Mr. Clark then presented to us the "new" distance formula
Below is the points and lines distance formula
distance=ax₁+by₁+c/√a²+b²

i.e) Find the distance from Point (2, -5) and the line 3x+4y+9=0
distance=ax₁+by₁+c/√a²+b²
=3(-2)+4(-5)+9/√3²+4²
=35/5
=7

HINT: Keep the Pythagurian Triples in mind when solving these equations
a²+b²=c²
*3, 4, 5
*6, 8, 10
*5, 12, 13
*12, 16, 20
*8, 15, 17
7, 24, 25

So I guess homework for the weekend is all uncompleted exercises up until..
Exercise 22: Distances between Points and Lines

Have a good weekend and enjoy!

Scribe for Monday will be Lauressa
since you expressed your ready-ness to blog!

P.S. Images will be placed tomorrow if I find the energy, sorry.