Let A be the set of all points ([], []) such that the area of triangle formed by the point…

Question

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

Accepted Answer

$$\left| {{1 \over 2}\left| {\begin{pmatrix} 5 & 6 & 1  \
   3 & 2 & 1  \
   \alpha  & \beta  & 1  \

\end{pmatrix} } \right|} \right| = 12$$

4$$\alpha$$ - 2$$\beta$$ = $$\pm$$ 24 + 8

$$\Rightarrow$$ 4$$\alpha$$ - 2$$\beta$$ = + 24 + 8 $$\Rightarrow$$ 2$$\alpha$$ - $$\beta$$ = 16

2x - y - 16 = 0 ..... (1)

$$\Rightarrow$$ 4$$\alpha$$ - 2$$\beta$$ = - 24 + 8 $$\Rightarrow$$ 2$$\alpha$$ - $$\beta$$ = -8

2x - y + 8 = 0 ...... (2)

Perpendicular distance of (1) from (0, 0)

$$\left| {{{0 - 0 - 16} \over {\sqrt 5 }}} \right| = {{16} \over {\sqrt 5 }}$$

Perpendicular distance of (2) from (0, 0)

$$\left| {{{0 - 0 + 8} \over {\sqrt 5 }}} \right| = {8 \over {\sqrt 5 }}$$

Solutions · 1