Answer :

Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations.

2x – 3y + 1 = 0 …(i)


x + y – 2 = 0 …(ii)


Now, we find the point of intersection of eq. (i) and (ii)


Multiply the eq. (ii) by 2, we get


2x + 2y – 4 = 0 …(iii)


On subtracting eq. (iii) from (i), we get


2x – 3y + 1 – 2x – 2y + 4 = 0


-5y + 5 = 0


-5y = -5


y = 1


Putting the value of y in eq. (ii), we get


x + 1 – 2 = 0


x – 1 = 0


x = 1


Hence, the point of intersection P(x1, y1) is (1, 1)



The equation of a line parallel to y – axis is of the form


x = a where a is some constant


Given that this equation of the line passing through the point of intersection (1, 1)


Hence, point (1, 1) will satisfy the equation of a line.



Putting x = 1 in the equation y = b, we get


x = a


1 = a


or a = 1


Now, required equation of line is x = 1



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