Answer:
y = 2b/3
Step-by-step explanation:
The x-coordinate of the intersection point of Line B D and Line C E is at [tex]\frac{2(a+c)}{3}[/tex]. Given that:
[tex]y=\frac{b}{a-2c}x -\frac{2bc}{a-2c} \\\\The\ y\ coordinate\ can\ be \ gotten\ by\ substituting\ the \ value\ of\ x\ and\ simplifying.\\ Substituting\ x:\\\\y=\frac{b}{a-2c}(\frac{2(a+c)}{3} ) -\frac{2bc}{a-2c}[/tex]
[tex]Simplyfing\ the\ parenthesis\\y=\frac{2b(a+c)}{3(a-2c)} -\frac{2bc}{a-2c}\\\\y=\frac{2ab+2bc}{3(a-2c)} -\frac{2bc}{a-2c}\\\\Simplyfying\ using\ LCF\\y=\frac{2ab+2bc-6bc}{3(a-2c)}\\\\y=\frac{2ab-4bc}{3(a-2c)}\\\\Factorizing:\\\\y=\frac{2b(a-2c)}{3(a-2c)}\\\\y=\frac{2b}{3}[/tex]
The y-coordinate of the intersection point of Line B D and Line C E is at [tex]\frac{2b}{3}[/tex].
