Math Problem Help!?

Find the equation of line A:
x1,y1 = -3,-1
x2,y2 = 1, 11

slope = (y2-y1)/(x2-x1) = (11+1)/(1+3) = 12/4 = 3
Equation : y=3x+b
substitute x=1 , y=11
11 = 3(1) + b
b= 11-3 = 8

Equation of line A is y=3x+8

Find the equation of line B:
x1,y1 = 5,1
x2,y2 = -5,21

slope = (y2-y1)/(x2-x1) = (21-1)/(-5-5) = 20/(-10) =-2
Equation : y=-2x+b
substitute x=5 , y=1
1 = -2(5) + b
b= 1+10 = 11

Equation of line B is y=-2x+11


Equate line A and line B

3x+8 = -2x+11
5x=3
x=3/5 = 0.6

substitute x=3/5 into equation of line A
y = 3x+8
y = 3(3/5)+8
y= 9/5+8
y = 49/5

The point of intersection of A and B is (3/5, 49/5)

Let (x, y) the point of intersection.

Slope of Line A:
(y + 1)/(x + 3) = (11 + 1)/(1 + 3)
(y + 1)/(x + 3) = 3
y + 1 = 3x + 9
3x - y = -8 …… [1]

Slope of Line B:
(y - 1)/(x - 5) = (21 - 1)/ (-5 - 5)
(y - 1)/(x - 5) = -2
y - 1 = -2x + 10
2x + y = 11 …… [2]

[1] + [2]:
5x = 3
x = 0.6

Subsyitute [1] x =0.6 into [2]:
2(0.6) + y = 11
y = 9.8

The point of intersection = (0.6, 9.8)

The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept


How to get the equation of the line that passes through A (- 3 ; - 1) B (1 ; 11) ?

To calculate m

m = (yB - yA) / (xB - xA) = (11 + 1) / (1 + 3) = 12/4 = 3

The equation of the line (AB) becomes: y = 3x + b


To calculate b

The line (AB) passes through A, so the coordinates of this point must verify the equation of the line.

y = 3x + b

b = y - 3x → you substitute x and y by the coordinates of the point A (- 3 ; - 1)

b = - 1 - (3 * - 3)

b = 8

→ The equation of the line (AB) is: y = 3x + 8


How to get the equation of the line that passes through C (5 ; 1) D (- 5 ; 21) ?

To calculate m

m = (yB - yA) / (xB - xA) = (21 - 1) / (- 5 - 5) = 20/- 10 = - 2

The equation of the line (CD) becomes: y = - 2x + b


To calculate b

The line (CD) passes through C, so the coordinates of this point must verify the equation of the line.

y = - 2x + b

b = y + 2x → you substitute x and y by the coordinates of the point C (5 ; 1)

b = 1 + (2 * 5)

b = 11

→ The equation of the line (CD) is: y = - 2x + 11


y = 3x + 8 ← this is the line (AB)

y = - 2x + 11 ← this is the line (CD)


Intersection between the line (AB) and the line (CD).

y = y

3x + 8 = - 2x + 11

3x + 2x = 11 - 8

5x = 3

→ x = 3/5


Recall the line (AB): y = 3x + 8

y = [3 * (3/5)] + 8

y = (9/5) + (40/5)

→ y = 49/5

The coordinates of the point of intersection are (3/5 ; 49/5)

The intersection of the line through (x₀, y₀) and (x₁, y₁) and the line through (x₂, y₂) and (x₃, y₃) (where the lines are not parallel) is:
( ((y₀x₁-x₀y₁)(x₃-x₂)+(x₀-x₁)(y₂x₃-x₂y₃)) / ((y₀-y₁)(x₃-x₂)+(x₀-x₁)(y₂-y₃)),
((y₀-y₁)(y₂x₃-x₂y₃)+(x₀y₁-y₀x₁)(y₂-y₃)) / ((y₀-y₁)(x₃-x₂)+(x₀-x₁)(y₂-y₃)) )

or, you can determine the formulas of each line, and then determine their intersection.

slope of A = (11 - (-1))/(1 - (-3))

= 12/4

= 3

equation of A

y + 1 = 3(x + 3)

y = 3x + 9 - 1

= 3x + 8 ...........(1)

slope of B = (21 - 1)/(-5 - 5)

= 20/-10

= - 2

equation of B

y - 1 = -2(x - 5)

y = -2x + 10 + 1

= - 2x + 11 ............(2)

equate (1) and (2)

3x + 8 = - 2x + 11

5x = 3

x = 3/5

Put in (1)

y = 3(3/5) + 8

= 9/5 + 8

= 9 4/5

Point of intersection = ( 3/5, 9 4/5).........[= (0.6, 9.8)]

A has slope 3, so, y = 3x + c
11 = 3 + c, so it was y = 3x + 8
B has slope – 2, so, y = -2x + k
1 = -2*5 + k, so, y = -2x + 11

3x + 8 = -2x + 11 at intersection,
so, x = 3/5, y = 9/5 + 8 = 49/5

Line A
m = 12 / 4 = 3
y + 1 = 3 ( x + 3 )
y + 1 = 3x + 9
3x - y = - 8

Line 2
m = 20 / (-10) = (-2)
y - 1 = (-2) ( x - 5 )
y = (-2) x + 11
2x + y = 11

3x - y = - 8
2x + y = 11_____add

5x = 3
x = 3/5

6/5 + y = 11
y = 49/5

Point of intersection is ( 3/5 , 49/5 )

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