Given equations of lines are x + y – 2 = 0 ...(i)and 2x – 3y + 4 = 0 ...(ii)Multiplying equation (i) by 3, we get3x + 3y – 6 = 0 ...(iii)Adding equation (ii) and (iii), we get 5x – 2 = 0 ∴ x = `2/5` Substituting x = `2/5` in equation (i), we get `2/5 + y - 2` = 0 ∴ y = `2 - 2/5 = 8/5` ∴ The required line passes through point `(2/5, 8/5)`. Also, the line makes intercept of 3 on X-axis∴ it also passes through point (3, 0). ∴ required equation of line passing through points `(2/5, 8/5)` and (3, 0) is `(y - 8/5)/(0 - 8/5) = (x - 2/5)/(3 - 2/5)` ∴ `((5y - 8)/5)/(-8/5) = ((5x - 2)/5)/(13/5)` ∴ `(5y - 8)/(-8) = (5x - 2)/13` ∴ 13 (5y – 8) = – 8 (5x – 2)∴ 65y – 104 = – 40x + 16∴ 40x + 65y – 120 = 0 ∴ 8x + 13y – 24 = 0 which is the equation of the required line.
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