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Mathematics Question on Angle between a Line and a Plane

LetA(1,1),B(−4,3)A(1, 1), B(-4, 3)and C(−2,−5)C(-2, -5) be vertices of a triangle ABC, P be a point on side BC, and Δ1Δ1 and Δ2Δ2 be the areas of triangles APB and ABC, respectively. If Δ1:Δ2=4:7Δ1 : Δ2 = 4 : 7, then the area enclosed by the lines AP, AC and the x-axis is

A

14\frac{1}{4}

B

34\frac{3}{4}

C

12\frac{1}{2}

D

1

Answer

12\frac{1}{2}

Explanation

Solution

The correct answer is (C):
Δ1Δ2=12×BP×AH12×BC×AH\frac{\Delta_1}{\Delta_2} = \frac{\frac{1}{2} \times BP \times AH}{\frac{1}{2} \times BC \times AH}
=47=\frac{ 4}{7}
vertices of a triangle ABC, P be a point on side BC
P(−207,−117)P\left(-\frac{20}{7}, -\frac{11}{7}\right)
Line AC:y–1=2(x–1)AC : y – 1 = 2(x – 1)
Intersection with x -axis
=(12,0)= (\frac{1}{2}, 0)
Line AP:y−1AP: y-1
=23(x−1)=\frac{ 2}{3}(x-1)
Intersection with x -axis
(−12,0)(\frac{-1}{2}, 0)
Vertices are (1,1),(12,0),(−12,0)(1,1), \left(\frac{1}{2},0\right), \left(-\frac{1}{2},0\right)
Area = 12 sq.unit\frac{1}{2}\text{ sq.unit}