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Question
My problems (set 1):
5. Find the slope and y-intercept of:
(e) (x/a)+(y/b)=1
17. The line is the perpendicular bisector of the line segment joining (2,8) and (-4,6).
27. In each, classify the lines as parallel, perpendicular, or neither.
(d) (Ax)+(By)+C=0 and (Bx)-(Ay)+D=0
29. Find the area of the triangle formed by the coordinate axes and the line through (1,4) and (2,1).
31. In each part, find the point of intersection of the lines.
(a) (2x)+(3y)=5 and y=(-1)
(b) (4x)+(3y)=(-2) and (5x)-(2y)=9

Answer
5.)
e.)
(x/a) + (y/b) = 1
xb + ay = ab
ay = -xb + ab
y = (-b/a)x + b

slope is (-b/a)
y-intercept is b

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17.)
(2,8) and (-4,6)
m = (6 - 8)/(-4 - 2)
m = (-2)/(-6)
m = (1/3)

(2,8), m = -3
8 = -3(2) + b
b = 14

y = -3x + 14

Slope = -3
y-intercept is 14

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27.)
Ax + By + C = 0
Bx - Ay + D = 0

By = -Ax - C
y = (-a/b)x - (c/b)

-Ay = -Bx - D
y = (b/a) + (d/a)

the lines are perpendicular

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29.)
(1,4) and (2,1)
m = (1 - 4)/(2 - 1)
m = -3

(1,4), m = -3
4 = -3(1) + b
b = 7

y = -3x + 7

now all you have to do is find the x intercept

0 = -3x + 7
-7 = -3x
x = (7/3)

so now you have (0,0), ((7/3),0) and (0,7)

D = 7/3
D = 7

D = sqrt((49/9) + 49)
D = sqrt(490/9)
D = sqrt((49/9) * 10)
D = (7/3)sqrt(10)

so now you now it's a scalene triangle

s = (a + b + c)/2
K = sqrt[s(s - a)(s - b)(s - c)]

s = ((7/3)sqrt(10) + 7 + (7/3))/2
s = ((7sqrt(10) + 21 + 7)/3)/2
s = (7sqrt(10) + 28)/6

K = sqrt(((7sqrt(10) + 28)/6)(((7sqrt(10) + 28)/6) - (7/3))(((7sqrt(10) + 28)/6) - 7)(((7sqrt(10) + 28)/6) - (7/3)sqrt(10)))

K = sqrt(((7sqrt(10) + 28)/6)(((7sqrt(10) + 28 - 14)/6)((7sqrt(10) + 28 - 42)/6)((7sqrt(10) + 28 - 42sqrt(10))/6)))

K = sqrt(((7sqrt(10) + 28)/6)((7sqrt(10) + 14)/6)((7sqrt(10) - 14)/6)((35sqrt(10) + 28)/6))

K = about 39.7148

i used www.quickmath.com to do it the quick way.

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31.)
a.)
2x + 3y = 5
y = -1

2x + 3(-1) = 5
2x - 3 = 5
2x = 8
x = 4

ANS : (4,-1)

b.)
4x + 3y = -2
5x - 2y = 9

multiply top by 2 and bottom by 3

8x + 6y = -4
15x - 6y = 27

23x = 23
x = 1

4(1) + 3y = -2
3y = -6
y = -2

ANS : (1,-2)

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