CAT 1996 — QA Question 29

The three lines can be expressed as $Y=\frac{5}{3}-\frac{2X}{3},Y=\frac{5X}{7}+\frac{2}{7}$ and $\frac{9X}{5}-\frac{4}{5}.$ Therefore, the slopes of the three lines are $\frac{-2}{3},\frac{5}{7}$ and $\frac{9}{5}$ respectively. For any two lines to be perpendicular to each other, the product of their slopes = –1. We find that the product of none of the slopes is –1. For any two be parallel, their slopes should be the same. This is again not the case. And finally for the two lines to be intersecting at the same point, there should be one set of values of (X, Y) that should satisfy the equations of 3 lines. Solving the first two equations, we get X = 1 and Y = 1. If we substitute this in the third equation, we find that it also satisfies that equation. So the solution set (1, 1) satisfies all three equations, suggesting that the three lines intersect at the same point, viz. (1, 1). Hence, they are coincident.