Help me! Maths?

1.
Equation of the line :
(y - 1)/(x - 7) = -1
y - 1 = -x + 7
x + y - 8 = 0


2.
Equation of the line :
[y + (2/5)]/[x - (1/4)] = 7/3
7[x - (1/4)] = 3[y + (2/5)]
7x - (7/4) = 3y + (6/5)
20[7x - (7/4)] = 20[3y + (6/5)]
140x - 35 = 60y + 24
140x - 60y - 59 = 0


3.
Equation of the line :
(y - 1)/(x - 2) = 1/2
x - 2 = 2(y - 1)
x - 2 = 2y - 2
x - 2y = 0


4.
When the question is : (5/7, 5) and ( -5, 4) on line

Equation of the line :
(y - 4)/(x + 5) = (5 - 4)/[(5/7) + 5]
(y - 4)/(x + 5) = 1/(40/7)
(y - 4)/(x + 5) = 7/40
7(x + 5) = 40(y - 4)
7x + 35 = 40y - 160
7x - 40y + 195 = 0


5.
[y + (1/7)]/[x + (1/2)] = [1 + (1/7)]/[(2/3) + (1/2)]
[y + (1/7)]/[x + (1/2)] = (8/7)/(7/6)
[y + (1/7)]/[x + (1/2)] = 48/49
48[x + (1/2)] = 49[y + (1/7)]
48x + 24 = 49y + 7
48x - 49y + 17 = 0


6.
Since the y-coordinates of the two points are both equal to 0, then the equation of the line is :
y = 0


7.
The general formula of horizontal line is : y = constant
Then equation of the required line is : y = 9