The equation of a straight line passing through the point \((4,3)\) and making intercepts on the coordinate axes whose sum is \(-1\) is given by (a) \(3 \mathrm{x}-2 \mathrm{y}=6\) and \(\mathrm{x}-2 \mathrm{y}=-2\) (b) \(3 x-2 y=-6\) and \(x-2 y=2\) (c) \(3 x-2 y=6\) and \(x+2 y=2\) (d) \(3 x-2 y=-6\) and \(x-2 y=-2\)

First, we will check whether the point (4,3) lies on the line \(3x-2y=6\) and \(x-2y=-2\). Plug in the coordinates \((x, y) = (4, 3)\): For the first equation: \( 3(4) - 2(3) = 12 - 6 = 6 \); The point lies on this line because the equation holds true. For the second equation: \( 4 - 2(3) = 4 - 6 = -2 \); The point lies on this line as well because the equation holds true. Now we need to check if the sum of the intercepts on coordinate axes is -1. For the first equation: When x=0, y=-3. So, the y-intercept is -3. When y=0, x=2. So, the x-intercept is 2. x-intercept + y-intercept = 2 - 3 = -1 For the second equation: When x=0, y=1. So, the y-intercept is 1. When y=0, x=-2. So, the x-intercept is -2. x-intercept + y-intercept = -2 + 1 = -1 Both conditions hold, so option (a) is correct.

We will now perform the same verification for option (b), but we already found the correct answer in option (a), so it is unnecessary to check the rest. However, let's verify one equation in option (b) to reinforce the conclusion. For equation \(3x - 2y = -6\) with point (4,3): \(3(4) - 2(3) = 12 - 6 = 6\); Since the equation doesn't hold true for point (4,3), we can conclude that option (b) is incorrect. In similar fashion, we could check options (c) and (d) – however, as we already found the correct answer, it is not required to evaluate further.