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Given:

2x – 7y + 11 = 0 and x + 3y – 8 = 0

To find:

The equation of the line passing through the point of intersection of 2x – 7y + 11 = 0 and x + 3y – 8 = 0 and is parallel to (i) x = axis (ii) y-axis.

Explanation:

The equation of the straight line passing through the points of intersection of 2x 7y + 11 = 0 and x + 3y 8 = 0 is given below:

2x 7y + 11 + λ(x + 3y 8) = 0

(2 + λ)x + ( 7 + 3λ)y + 11 8λ = 0
(i) The required line is parallel to the x-axis. So, the coefficient of x should be zero.

2 + λ = 0

λ = -2

Hence, the equation of the required line is

0 + ( 7 6)y + 11 + 16 = 0

13y 27 = 0

(ii) The required line is parallel to the y-axis. So, the coefficient of y should be zero.

-7 + 3λ = 0

λ Hence, the equation of the required line is 13x – 23 = 0

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