CAT 2006 — QA Question 11

All the given graphs are drawn to the same scale.

We can see that the line makes an angle which is more than 45° with the horizontal axis.

∴ The slope of the line is greater than 1.

Let the slope be k.

∴ (y – x) = k(y + x)     {∵ k > 1}

∴ y – x = ky + kx

$\therefore y=\frac{x(k+1)}{(1-k)}$

$\frac{(k+1)}{(1-k)}$ is negative and $\left|\frac{(k+1)}{(1-k)}\right|>1$

∴ The graph of y against x will be such that when x is positive, y is negative and |x| < |y|, except at (0, 0).

Hence, option (d).