Question

Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

Answer 1

slope of line parallel to Y-axis is 1/0 

Here , equation of lines are 
x - 7y + 5 = 0------------(1)
3x + y = 0 ----------------(2)

solve equations (1) and (2) for finding point of intersection of these two lines .

x - 7y + 5 + 21x + 7y = 0
22x + 5 = 0
x = -5/22 , put it in equation (2) 
y = -3x = 15/22 
hence, the point is ( -5/22, 15/22) 

Equation of required line by using 
y - y1 = m(x - x1) 
(y - 15/22) = 1/0(x + 5/22)
0 = x + 5/22
22x + 5 = 0 

Therefore, required line: 22x + 5 = 0 

Answer 2

Given Equations of lines are 

x - 7y + 5 = 0  ---- (1)

3x + y = 0

y = -3x. ---- (2)

Substitute y in (1), we get

x + 21x + 5 = 0

22x + 5 = 0

22x = -5

x = -5/22.

Substitute x in (2), we get

y = -3(-5/22)

y = 15/22.

Now (x,y) = (-5/22,15/22).

Given that Equation of line parallel to the y-axis is x = a.

x = -5/22

22x = -5.

22x + 5 = 0.


Hope this helps!