Straight Lines
Straight Line
Allen Star Batch
Grade 11
Question:
Given $\triangle ABC$ whose vertices are $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$. Let there exists a point $P(a, b)$ such that $6a = 2x_1 + x_2 + 3x_3; 6b = 2y_1 + y_2 + 3y_3$
$P(a,b)$ lies inside the $\triangle ABC$
Area of triangle $PBC$ is less than area of $\triangle ABC$
$P(a,b)$ lies outside the $\triangle ABC$
Area of triangle $PBC$ is greater than area of $\triangle ABC$
Step-by-Step Solution
Key Concept: A point inside a triangle has coordinates as a convex combination of vertices, and such regions partition the original triangle.
Point $D$ divides $AB$ in ratio $1:2$, so $D = \left(\frac{2x_1 + x_2}{3}, \frac{2y_1 + y_2}{3}\right)$. Point $P$ is the midpoint of $DC$, giving $P = \left(\frac{2x_1 + x_2 + 3x_3}{6}, \frac{2y_1 + y_2 + 3y_3}{6}\right)$. Since $P$ lies inside triangle $ABC$ (as a weighted average of vertices with positive coefficients), the area of triangle $APBC$ is less than the area of triangle $ABC$.
Correct Answer: 1,2