Question

Let\( A(1, 1), B(-4, 3) \)and \(C(-2, -5)\) be vertices of a triangle ABC, P be a point on side BC, and \(Δ1\) and \(Δ2\) be the areas of triangles APB and ABC, respectively. If \(Δ1 : Δ2 = 4 : 7\), then the area enclosed by the lines AP, AC and the x-axis is

• A. \(\frac{1}{4}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{1}{2}\)
• D. 1

Answer 1

The Correct Option is C

The correct answer is (C):
\(\frac{\Delta_1}{\Delta_2} = \frac{\frac{1}{2} \times BP \times AH}{\frac{1}{2} \times BC \times AH}\)
\(=\frac{ 4}{7}\)

\(P\left(-\frac{20}{7}, -\frac{11}{7}\right)\)
Line \(AC : y – 1 = 2(x – 1)\)
Intersection with x-axis
\(= (\frac{1}{2}, 0)\)
Line \(AP: y-1\)
\(=\frac{ 2}{3}(x-1)\)
Intersection with x-axis
\((\frac{-1}{2}, 0)\)
Vertices are \((1,1), \left(\frac{1}{2},0\right), \left(-\frac{1}{2},0\right)\)
Area = \(\frac{1}{2}\text{ sq.unit}\)