Let A(1,1),B(−4,3),C(−2,−5) be vertices of a triangle ABC,P be a point on side BC, and \[\Delta_{1}\] and \[\Delta_{2}\] be the areas of triangles APB and ABC, respectively. If \[\Delta_{1}: \Delta_{2}=4: 7\], then the area enclosed by the lines AP,AC and the x-axis is :
Solutions · 1
Official solution
\[{{{\Delta _1}} \over {{\Delta _2}}} = {{{1 \over 2} \times BP \times AH} \over {{1 \over 2} \times BC \times AH}} = {4 \over 7}\]
\[P\left( {{{ - 20} \over 7},{{ - 11} \over 7}} \right)\]
Line AC:y−1=2(x−1)
Intersection with x-axis \[ = \left( {{1 \over 2},0} \right)\]
Line \[AP:y - 1 = {2 \over 3}(x - 1)\]
Intersection with x-axis \[\left( {{{ - 1} \over 2},0} \right)\]
Vertices are \[(1,1),\left( {{1 \over 2},0} \right)\] and \[\left( {{{ - 1} \over 2},0} \right)\]
Area \[ = {1 \over 2}\] sq. unit