
Topic review (newest first)
 MathsIsFun
 20051122 08:58:01
At (2,4): 4 = m(2) + b → b = 4+2m At (x1,0): 0 = m(x1) + b → x1 = b/m At (0,y1) y1 = m(0) + b → y1 = b
And we are also given: x1+y1 = 4
So we have four equations, and four unknowns (m,b,x1,y1), so now we can hopefully just use algebra to solve it.
Start with: x1+y1 = 4 Substitute for x1 and y1: b/m + b = 4 Multiply by m: b + bm = 4m Put at one side: 4m + bm  b = 0 Replace b with (4+2m): 4m +(4+2m)m  (4+2m) = 0 Expand: 4m + 4m + 2m²  4  2m = 0 Collect terms: 2m² + 6m  4 = 0
Now just solve the quadratic equation here and you get:
m = 0.56 and 3.56 Actual values are (6±√68)/4
NOW, let us check the answer (just the 0.56 one):
b = 4+2m = 5.12 At x=2, y=0.56(2)+5.12 = 4 At x=0, y1 = 0.56(0)+5.12 = 5.12 At y=0, 0 = 0.56(x1)+5.12, hence x1 = 5.12/0.56 = 9.14
x1+y1 = 5.129.14 = 4.02 (close enough to 4), so that looks right!
If this doesn't agree with the answer you have, then ... who knows!
 Sephiroth Valentine
 20051122 04:19:45
At (2,4): 4 = m(2) + c → c = 4+2m At (x1,0): 0 = m(x1) + c→ x1 = c/m At (0,y1) y1 = m(0) + c → y1 = c
x1+y1 = 4 ∴c/m +c=4/1 y1=c c=2m+4
next substitute c value for c into equation: c/m +c=4/1
(2m+4)/m +2m+4=4
2m+4+2m²+4m=0 since 2m+4=c==y1 then x1+4= 2m²+4m
I still can't get it out any ideas anyone. I guess I went hopelessly wrong
 MathsIsFun
 20051121 16:59:32
Have a go ... copy and paste what I did and make the needed changes  I think you will still end up with a quadratic equation. Post your answer here so I can see.
 Sephiroth Valentine
 20051121 08:48:05
Sorry I see what was wrong I made a mistake when typing the question. I should have been x1 + y1=4 not x1 + y1=4m . Thanks for all your help though. Can your method still work with this equation?
 Sephiroth Valentine
 20051121 08:45:08
Something seems worng as I checked the answer at the back of the book and the answers for m were supposed to be either 1 or 2
 MathsIsFun
 20051121 08:14:33
I think it is just a bunch of formulas to solve.
Using the Line Equation: y = mx+b
At (2,4): 4 = m(2) + b → b = 4+2m At (x1,0): 0 = m(x1) + b → x1 = b/m At (0,y1) y1 = m(0) + b → y1 = b
And we are also given: x1+y1 = 4m
So we have four equations, and four unknowns (m,b,x1,y1), so now we can hopefully just use algebra to solve it.
There may be a quicker way to do this, but I am just going to start anywhere and see what happens:
Start with: x1+y1 = 4m Substitute for x1 and y1: b/m + b = 4m Looks "quadratic", so multiply by m: b + bm = 4m² Put in quadratic form: 4m² +bm  b = 0 Replace b with (4+2m): 4m² +(4+2m)m  (4+2m) = 0 Expand: 4m² + 4m + 2m²  4  2m = 0 Collect terms: 6m² + 2m  4 = 0
Now just solve the quadratic equation here and you get:
m = 2/3 or 1
I will leave you to check that this actually works (my error rate is about 1 in 10 )
 Sephiroth Valentine
 20051121 06:45:58
6(C) A line containg the point(2,4) has a slope of m which is not equal to zero. The line intercepts the xaxis at (x1,0) and the y axis at (0,y1) If x1 + y1=4m find the slopes of the two lines that satisfy this condition. Hence find the tangent of the acute angle between these two lines.
I really don't know where to start Tried using the slope formula y2y1/x2x1 using the point (2,4) and (x1,0) and then I used the points (2,4) and (0,y1) to get another value for m. I put bothe equal to one another but I could'nt get a value to work out. Could any help me please.
