Question:
For some \(\alpha \in \mathbb{N}\), let \(\mathrm{PQR}\) be a triangle with two fixed vertices \(\mathrm{P}(2, 5)\) and \(\mathrm{Q}(\alpha, -11)\). If the point \(\mathrm{R}\) moves on the line \(l_1 : 9x + 7y + \alpha = 0\), then the centroid of \(\Delta \mathrm{PQR}\) moves on the line \(l_2\), which is parallel to \(l_1\) at a distance \(\frac{20}{3\sqrt{130}}\) units from it. If the distance of \(\mathrm{Q}\) from \(l_2\) is \(\frac{k}{3\sqrt{130}}\), then k is equal to :
For some \(\alpha \in \mathbb{N}\), let \(\mathrm{PQR}\) be a triangle with two fixed vertices \(\mathrm{P}(2, 5)\) and \(\mathrm{Q}(\alpha, -11)\). If the point \(\mathrm{R}\) moves on the line \(l_1 : 9x + 7y + \alpha = 0\), then the centroid of \(\Delta \mathrm{PQR}\) moves on the line \(l_2\), which is parallel to \(l_1\) at a distance \(\frac{20}{3\sqrt{130}}\) units from it. If the distance of \(\mathrm{Q}\) from \(l_2\) is \(\frac{k}{3\sqrt{130}}\), then k is equal to :
Show Hint
Find the locus of the centroid, use the given distance between \(l_1\) and \(l_2\) to fix \(\alpha\), then compute the distance of Q from \(l_2\).
Updated On: Nov 12, 2026
- 117
- 129
- 131
- 133
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The Correct Option is C
Solution and Explanation
Let R = (h, m) on \(l_1\) and the centroid G = (X, Y). Then \(X = \dfrac{2 + \alpha + h}{3}\), \(Y = \dfrac{5 - 11 + m}{3}\), so \(h = 3X - 2 - \alpha\), \(m = 3Y + 6\).
Substituting in \(9h + 7m + \alpha = 0\): \(27X - 18 - 9\alpha + 21Y + 42 + \alpha = 0\), i.e. \(27X + 21Y + 24 - 8\alpha = 0\), i.e. \(l_2 : 9x + 7y + \dfrac{24 - 8\alpha}{3} = 0\).
Distance between \(l_1\) and \(l_2\) = \(\dfrac{|\alpha - (24 - 8\alpha)/3|}{\sqrt{130}} = \dfrac{|11\alpha - 24|}{3\sqrt{130}}\). Setting this equal to \(\dfrac{20}{3\sqrt{130}}\): \(11\alpha - 24 = \pm 20\), so \(\alpha = 4\) (the other value, 4/11, is not a natural number).
Then \(l_2 : 9x + 7y - \dfrac{8}{3} = 0\). For Q(4, -11): \(9(4) + 7(-11) - \dfrac{8}{3} = -41 - \dfrac{8}{3} = -\dfrac{131}{3}\).
Distance = \(\dfrac{131/3}{\sqrt{130}} = \dfrac{131}{3\sqrt{130}}\), so k = 131.
Answer: (3) 131
Substituting in \(9h + 7m + \alpha = 0\): \(27X - 18 - 9\alpha + 21Y + 42 + \alpha = 0\), i.e. \(27X + 21Y + 24 - 8\alpha = 0\), i.e. \(l_2 : 9x + 7y + \dfrac{24 - 8\alpha}{3} = 0\).
Distance between \(l_1\) and \(l_2\) = \(\dfrac{|\alpha - (24 - 8\alpha)/3|}{\sqrt{130}} = \dfrac{|11\alpha - 24|}{3\sqrt{130}}\). Setting this equal to \(\dfrac{20}{3\sqrt{130}}\): \(11\alpha - 24 = \pm 20\), so \(\alpha = 4\) (the other value, 4/11, is not a natural number).
Then \(l_2 : 9x + 7y - \dfrac{8}{3} = 0\). For Q(4, -11): \(9(4) + 7(-11) - \dfrac{8}{3} = -41 - \dfrac{8}{3} = -\dfrac{131}{3}\).
Distance = \(\dfrac{131/3}{\sqrt{130}} = \dfrac{131}{3\sqrt{130}}\), so k = 131.
Answer: (3) 131
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