Given that 16p^{2}+49q^{2}-4r^{2}-56pq=0. Which one of the following is a point on a pair of straight lines (px+qy+r)(px+qy-r)=0?

Explanation

The pair of lines (px+qy+r)(px+qy-r)=0 expands to (px+qy)^2 - r^2 = 0, or (px+qy)^2 = r^2. From the given condition 16p^2+49q^2-56pq = 4r^2, we can rewrite it as (4p-7q)^2 = 4r^2, which simplifies to r^2 = (2p - \frac{7}{2}q)^2. Substituting this back gives (px+qy)^2 = (2p - \frac{7}{2}q)^2. By comparing terms, the point (x,y) that satisfies this is x=2 and y=-\frac{7}{2}.

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