Çözüldü Coordinate Geometry - Slope of Line - Quadratic Equations - Triangle Area

If the sum of the slopes of two lines that vertically cross on point A(3, 4) is 3 / 2, and their x-intercepts are points B and C, what is the value of this ABC triangle area?

Original Turkish wording of the question: https://i72.servimg.com/u/f72/19/97/10/39/sat21.png
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Let the slopes of the lines be p and q, and the two points on the line y = 0 (x-axis) be B(b, 0) and C(c, 0).
From the equations p·q = -1 and p + q = 3 / 2, 2p^2 - 3p - 2 = 0 ⇒ p1 = 2 ⇒ q1 = -1 / 2 and p2 = -1 / 2 ⇒ q2 = 2
Assuming slopes to be p = 2 and q = -1 / 2, equations of the lines passing through A(3, 4) are;
y - 4 = 2(x - 3) passing through B(b, 0) the x-intercept is 0 - 4 = 2(b - 3) ⇒ b = 1
y - 4 = (-1 / 2)·(x - 3) passing through C(c, 0) the x-intercept is 0 - 4 = (-1 / 2)·(c - 3) ⇒ c = 11
Area(ΔABC) = (11 - 1)·4 / 2 = 20 squared unit.

Notes:
1.
Or for the vertices A(3, 4), B(1, 0), C(11, 0) of the triangle ABC,
Area(ΔABC) = (1 / 2)·|x·(y - y) + x·(y - y) + x·(y - y)|
Area(ΔABC) = (1 / 2)·|x1·(y - y) + x2·(y - y) + x3·(y - y)|
Area(ΔABC) = (1 / 2)·|x1·(y2 - y3) + x2·(y3 - y1) + x3·(y1 - y2)|
Area(ΔABC) = (1 / 2)·|3·(0 - 0) + 1·(0 - 4) + 11·(4 - 0)| = 20 squared unit.

2.
I do not understand how such stupid problems remain unsolved in a Facebook Community of Turkey's Mathematics Teachers!