Let A be the set of all points (alpha, beta) | vedclass

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(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_

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$\frac{8}{\sqrt{5}}$

Vedclass · Answer

(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 12.$Substituting the given points $(5, 6), (3, 2),$ and $(\alpha, \beta)$:$\frac{1}{2} |5(2 - \beta) + 3(\beta - 6) + \alpha(6 - 2)| = 12$$|10 - 5\beta + 3\beta - 18 + 4\alpha| = 24$$|4\alpha - 2\beta - 8| = 24$$|2\alpha - \beta - 4| = 12$This gives two possible lines for the locus of point $A$:Case $1$: $2\alpha - \beta - 4 = 12 \Rightarrow 2\alpha - \beta - 16 = 0$Case $2$: $2\alpha - \beta - 4 = -12 \Rightarrow 2\alpha - \beta + 8 = 0$The distance from the origin $(0, 0)$ to a line $ax + by + c = 0$ is $\frac{|c|}{\sqrt{a^2 + b^2}}.$For line $1$: $d_1 = \frac{|-16|}{\sqrt{2^2 + (-1)^2}} = \frac{16}{\sqrt{5}}$For line $2$: $d_2 = \frac{|8|}{\sqrt{2^2 + (-1)^2}} = \frac{8}{\sqrt{5}}$The least distance is $\frac{8}{\sqrt{5}}.$

Let $A$ be the set of all points $(\alpha, \beta)$ such that the area of the triangle formed by the points $(5, 6), (3, 2),$ and $(\alpha, \beta)$ is $12 \text{ square units}.$ Then the least possible length of a line segment joining the origin to a point in $A$ is:

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