Given that 16p2 + 49q2 - 4r2 - 56pq = 0. Which one of the following is a point on a pair of straight lines (px + qy + r) (px + qy - r) = 0?
Step 1 — Factor the given condition.
\(16p^{2} - 56pq + 49q^{2} = (4p - 7q)^{2}\) and \(4r^{2} = (2r)^{2}\), so the condition \(16p^{2} + 49q^{2} - 4r^{2} - 56pq = 0\) is equivalent to:
\((4p - 7q)^{2} = (2r)^{2}\).
Step 2 — Rewrite the pair of lines.
\((px + qy + r)(px + qy - r) = 0\) is a difference of squares, equivalent to \((px + qy)^{2} = r^{2}\).
Step 3 — Test option (2, −7/2).
Substituting \(x = 2,\ y = -\tfrac{7}{2}\):
\(\left(2p - \tfrac{7}{2}q\right)^{2} = r^{2}\)
Multiplying both sides by 4:
\((4p - 7q)^{2} = 4r^{2}\) — exactly the condition from Step 1. ✓
So the point (2, −7/2) always lies on the pair of lines whenever the given condition on p, q, r holds.
Why the other options fail:
- (2, 7/2): gives \(16p^{2} + 56pq + 49q^{2} = 4r^{2}\) — opposite sign on the middle term; not equivalent to the condition.
- (4, −7): gives \((4p - 7q)^{2} = r^{2}\), not \(4r^{2}\); the LHS matches but the RHS is off by a factor of 4.
- (4, 7): gives \((4p + 7q)^{2} = r^{2}\) — wrong sign on the middle term and wrong factor on the RHS.
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