Q. 73.7( 3 Votes )

Find the equation

Answer :

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

x – 7y + 5 = 0 …(i)


3x + y – 7 = 0 …(ii)


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


Multiply the eq. (i) by 3, we get


3x – 21y + 15 = 0 …(iii)


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


3x + y – 7 – 3x + 21y – 15 = 0


22y - 22 = 0


22y = 22


y = 1


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


x – 7(1) + 5 = 0


x – 7 + 5 = 0


x – 2 = 0


x = 2


Hence, the point of intersection P(x1, y1) is (2, 1)



The equation of line parallel to x – axis is of the form


y = b where b is some constant


Given that this equation of the line passing through the point of intersection (2, 1)


Hence, point (2, 1) will satisfy the equation of a line.



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


y = b


1 = b


or b = 1


Now, the required equation of a line is y = 1



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