A rod of length 8 units moves such that its e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$23$

Vedclass · Answer

(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,$\alpha-\beta-\gamma = 13 - (-6) - (-4) = 13+6+4 = 23$.

Similar Questions

If $L : ax + by + c = 0$ is a variable straight line,where $a, b$ and $c$ are the second,fourth,and seventh terms of an $AP$ respectively,then $L$ passes through the fixed point:

If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ and $\frac{x}{\beta} + \frac{y}{\alpha} = 1$ meets the coordinate axes in $A$ and $B$,then the locus of the midpoint of $AB$ is

If $A(2, -3)$ and $B(-2, 1)$ are two vertices of a $\triangle ABC$ and if the centroid of $\triangle ABC$ lies on the line $2x + 3y = 1$,then the locus of vertex $C$ of $\triangle ABC$ is equal to

If $A$ is $(2, 5)$,$B$ is $(4, -11)$ and $C$ lies on $9x + 7y + 4 = 0$,then the locus of the centroid of the $\Delta ABC$ is a straight line parallel to the straight line:

If $a, b, c$ are in harmonical progression,then the line $bcx + cay + ab = 0$ passes through a fixed point whose coordinates are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module