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(B) Let the coordinates of $A$ be $(a, a+2)$ and $B$ be $(b, -2)$.Point $P(h, k)$ divides $AB$ in the ratio $2:1$.Using the section formula: $h = \fra

Vedclass · Accepted Answer

$23$

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(B) Let the coordinates of $A$ be $(a, a+2)$ and $B$ be $(b, -2)$.Point $P(h, k)$ divides $AB$ in the ratio $2:1$.Using the section formula: $h = \frac{2b+a}{3}$ and $k = \frac{2(-2)+1(a+2)}{3} = \frac{a-2}{3}$.From $k = \frac{a-2}{3}$,we get $a = 3k+2$.From $h = \frac{2b+a}{3}$,we get $2b = 3h-a = 3h-3k-2$,so $b = \frac{3h-3k-2}{2}$.The length of the rod $AB = 8$,so $AB^2 = 64$.$(b-a)^2 + (-2-(a+2))^2 = 64$$(\frac{3h-3k-2}{2} - (3k+2))^2 + (-4-a)^2 = 64$$(\frac{3h-9k-6}{2})^2 + (-4-(3k+2))^2 = 64$$\frac{9(h-3k-2)^2}{4} + (3k+6)^2 = 64$$9(h^2+9k^2+4-6hk-4h+12k) + 4(9k^2+36k+36) = 256$$9(h^2+9k^2-6hk-4h+12k+4+4k^2+16k+16) = 256$$9(h^2+13k^2-6hk-4h+28k+20) = 256$$9(h^2+13k^2-6hk-4h+28k) + 180 = 256$$9(h^2+13k^2-6hk-4h+28k) - 76 = 0$.Comparing with $9(x^2+\alpha y^2+\beta xy+\gamma x+28y)-76=0$,we get $\alpha=13$,$\beta=-6$,$\gamma=-4$.Therefore,$\alpha-\beta-\gamma = 13 - (-6) - (-4) = 13+6+4 = 23$.

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