Straight LineHardBloom L3
Question
Two sides of a triangle have the joint equation $(x - 3y + 2)(x + y - 2) = 0$, the third side which is variable always passes through the point $( - 5, - 1)$, then possible values of slope of third side such that origin is an interior point of triangle is/are:
Options
A.$\frac{- 4}{3}$
B.$\frac{- 2}{3}$
C.$\frac{- 1}{3}$
D.$\frac{1}{6}$
Solution
Sol. [Key Concept: Origin is interior of the triangle only for a slope range of the variable third side.]
$m \in \left( - 1,\frac{1}{5} \right)$ for origin to lie inside the triangle
✘ (a) $- \frac{4}{3}$ lies outside the admissible range $( - 1,\frac{1}{5})$.
✔ (b) $- \frac{2}{3}in( - 1,\frac{1}{5})$ — origin interior.
✔ (c) $- \frac{1}{3}in( - 1,\frac{1}{5})$ — origin interior.
✔ (d) $\frac{1}{6}in( - 1,\frac{1}{5})$ — origin interior.
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