Indian Statistical Institute ( ISI ) B.Math & B.Stat : Co-ordinate Geometry

Let $A$ be the set of all points $(h, k)$ such that the area of the triangle formed by $(h, k), (5, 6)$ and $(3, 2)$ is $12$ square units. What is the least possible length of a line segment joining $(0, 0)$ to a point in $A$? Take the base of the triangle to be the line segment obtained by joining the points $(5,6)$ and $(3,2)$. Equation of the base is $2x-y-4 =0$. Length of the base is $\sqrt{(5-3)^2+(6-2)^2} = 2 \sqrt{5}.$ Let $p$ be the length of the perpendicular from the point $(h,k)$ onto the base. ( Note that the point $(h,k)$ cannot lie on the base. Why?) Since the area is given to be $12$, $12 = \frac{1}{2} \times p \times 2 \sqrt{5} \implies p = \frac{12}{\sqrt{5}}$. Therefore the point $(h,k)$ lies at a distance of $\frac{12}{\sqrt{5}}$ units from the base on both sides. Thus $A$ is the set of all points on the line $parallel$ to the base and at a distance $\frac{12}{\sqrt{5}}$ units away from the base. In the diagram, the lines colored green represents the set $A$. Clearly the least possible length of a line segment joining $(0, 0)$ to a point in $A$? is the distance between the point $(0,0)$ and the line drawn parallel to the base and to the left side of the base. Let $XY$ be the line segment perpendicular to the base and the line and passing through the orgin as shown in the diagram. Required distance is $OX$ and $OX = XY - OY = \frac{12}{\sqrt{5}} - \big|\frac{-4}{\sqrt{2^2+(-1)^2}}\big| = \frac{12}{\sqrt{5}} - \frac{4}{\sqrt{5}} = \frac{8}{\sqrt{5}}$