Find X of point P
BC is perpendicular to AP; 4 times the area of triangle APC is 3 times the area of the triangle OBC
A=(-0.5;0) C=(1;0) B is on the y axis
The answer is 0.5.
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3 solutions
Frist part
Second part
Area of △ A P C = [ △ A P C ] = 2 3 ⋅ P y
Area of △ O B C = [ △ O B C ] = 1 ⋅ B y = B y
4 ⋅ [ △ A P C ] = 3 ⋅ [ △ O B C ]
4 ⋅ 2 3 ⋅ P y = 3 ⋅ B y
6 ⋅ P y = 3 ⋅ B y
B y = 2 ⋅ P y
The points P and B are on a straight line going through point C = ( 1 , 0 , therefore point P must be half way between B and C with an x -coordinate = 0 . 5 .
The problem would be more interesting if it were the y -coordinate of P that was asked for.
AP is tangent in P to the graphic of y ; if P=(Xo;Yo) I know the slope and so on ...