Hard maths question (1 Viewer)

[authenticity]

COORDINATE GEOMETRY
1. Explain why the gradient of a straight line is given by m in the equation y=mx+b
2. A straight line has equation y=mx+4m+3. What are the coordinates of a point through which this line passes through, no matter the value of m.
3. AO and BO represent two mirrors placed at right-angles. P is the light source. A ray of light, PQRS is reflected from AO at Q and from OB at R. What are the coordinates of R?
http://desmond.imageshack.us/Himg803/scaled.php?server=803&filename=22394985.jpg&res=landing
4. The line L₁ has equation y=mx+k. Line L₁ crosses the y-axis at P and line L₂ crosses the x-axis at Q. If PQ is perpendicular to both lines, then determine the y-intercept of L₂.
http://desmond.imageshack.us/Himg41/scaled.php?server=41&filename=28895550.jpg&res=landing
5. For which values of k do the equations kx-y=2 and x+y=3 has solution (x,y) in which x>0 and y>0?
A) k>-1 B) K<(2/3) C) -1<k<(2/3) D) K<-1 E) k>(2/3)
6. Suppose that A(x₁, y₁), B(x₂, y₂) and C(x₃,y₃) are collinear points in the real number plane, and T is the point (ux₁+vx₂+wx₃, uy₁+vy₂+wy₃), where u, v and w and non-zero and u+v+w=1.
A. show the x₁y₂+x₂y₃+x₃y₁ = x₁y₃+x₂y₁+x₃y₂.
B. Hence, show that T lies on the line through A, B and C.

Thank you so much if you work any of it out :')

 

[iBibah]

Why don't you show us what you have done/know then we can work from there instead of us doing it all?

 

[authenticity]

for Q. 3 All i've done is find the equation of PS. I don't know the angle of light reflection or anything... :S. for Q.5. I tried working it out with simultaneous equations, except it's not going anywhere.

 

[jxballistic]

2. A straight line has equation y=mx+4m+3. What are the coordinates of a point through which this line passes through, no matter the value of m.

y = (4+x)m +3
if x = -4, m will be eliminated
and y=3
therefore (-4,3)

 

[authenticity]

jxballistic, thanks for your help

 

[Bozza555]

i think this is it for question 4
<a href="http://www.codecogs.com/eqnedit.php?latex=Since \, the\, two\,lines\,are\,parallel\,\\\therefore m(L_2)=m(L_1)\\\therefore y-0=m(x-Q)\\y=m(x-Q)\\ at\,y\,intercept\,x=0\\y=m(0-Q)\\\therefore y=-mQ \\ Hence\,y\,intercept\,of\,L_2\,is (0,-mQ)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Since \, the\, two\,lines\,are\,parallel\,\\\therefore m(L_2)=m(L_1)\\\therefore y-0=m(x-Q)\\y=m(x-Q)\\ at\,y\,intercept\,x=0\\y=m(0-Q)\\\therefore y=-mQ \\ Hence\,y\,intercept\,of\,L_2\,is (0,-mQ)" title="Since \, the\, two\,lines\,are\,parallel\,\\\therefore m(L_2)=m(L_1)\\\therefore y-0=m(x-Q)\\y=m(x-Q)\\ at\,y\,intercept\,x=0\\y=m(0-Q)\\\therefore y=-mQ \\ Hence\,y\,intercept\,of\,L_2\,is (0,-mQ)" /></a>

 

[Bozza555]

couldn't be bothered to write full working out but let me know if you don't get any of it aha

 

[SpiralFlex]

*Editing

 

[SunnyScience]

1. Explain why the gradient of a straight line is given by m in the equation y=mx+b


Two points are needed to define a iline. Y = x defines a line at an angle of 45 degrees through the origin. In other words, when x=1, y=2..x=2,y=2 and so on.

m and b are now added to manipulated this line so all other possible linear lines can be defined.

Firstly, b represents the shift of this line up and down the y-axis. The line y=x can be moved up, say 1 unit, along the y-axis. This would define a completely new line (y=x+1). It can also be moved down the y-axis, say 2 units to define the line y=x-2. For generality, this shift up and down the y-axis is allocated the pro numeral b. This gives us the general equation of a line at 45 degress (y=x shifted up and down) i.e. y=x+b.

However, this still does not define all possible lines. Though we may have a line at 45 degrees at a point 1 unit above the x-axis (y=x+1), this line can be drawn though the point (x,1) infinitely. For example, a line parallel to the x-axis, or a really steep line almost pointing to infinite along the y-axis. To define a definite line, this change in slope must be accounted for. This is achieved by manipulating what happens to y when we input x-values. Going back to our starting line, y=x, if we put in x=1, we will get y=1. However, if we add a change in slope of 2, y will now equal twice the value of the x-point and will therefore be much more steep then say y=x, compared to y=2x. This change in slope is generalised with the pro numeral m. Dwelling deeper, this represents the change in vertical rise, over the horizontal run of the determined x and y points e.g. y=2/x=1 (2). When the y-intercept is re-introduced, this defines a unique line in Euclidean space that can only be expressed in through single equation. For example, y=5x+7 will only ever define a single line in flat space.

Therefore, without m (slope), an infinite amount of lines can be defined as the third variable is not taken into account (the points the line goes through, the shift up and down, and the slope), rather than the single line we are interested in.

Hope this is somewhat accurate and helpful

 

[Leffife]

@SunnyScience

You should start writing math textbooks.

You sure do enjoy writing a lot.

 

[SunnyScience]

@SunnyScience

You should start writing math textbooks.

You sure do enjoy writing a lot.

lol...when I see something that I actually might know, I go all out.
Though I do hope to tutor lower levels of maths at the end of the year, because I really do enjoy explaining it

 

[Triage]

Sinx + cosx = 0
Solve

 

[jxballistic]

Sinx + cosx = 0
Solve

Sinx = -Cosx
(sinx)/(cosx)=-1
tan x = -1
x = -45, AND -45+360 = 315