You are not logged in.

- Topics: Active | Unanswered

**Sephiroth Valentine****Member**- Registered: 2005-11-19
- Posts: 6

6(C) A line containg the point(-2,4) has a slope of m which is not equal to zero. The line intercepts the x-axis 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 y2-y1/x2-x1 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.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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 )

Offline

**Sephiroth Valentine****Member**- Registered: 2005-11-19
- Posts: 6

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

Offline

**Sephiroth Valentine****Member**- Registered: 2005-11-19
- Posts: 6

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?

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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.

Offline

**Sephiroth Valentine****Member**- Registered: 2005-11-19
- Posts: 6

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

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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 = 5.12-9.14 = -4.02 (close enough to -4), so that looks right!

If this doesn't agree with the answer you have, then ... who knows!

Offline