Question

The x-coordinate of the intersection point of Line B D and Line C E is StartFraction 2 (a + c) Over 3 EndFraction. y = (StartFraction b Over a minus c EndFraction)x − (StartFraction 2 b c Over a minus 2 c EndFraction) y = (StartFraction b Over a minus 2 c EndFraction) (StartFraction 2 (a + c) Over 3 EndFraction) minus (StartFraction 2 b c Over a minus 2 c EndFraction) y = (StartFraction b Over a minus 2 c EndFraction) (StartFraction 2 (a + c) Over 3 EndFraction) minus (StartFraction 6 b c Over 3(a minus 2 c) EndFraction) y = StartFraction 2 b (a + c) minus 6 b c Over 3 (a minus 2 c) EndFraction y = StartFraction 2 a b + 2 b c minus 6 b c Over 3 (a minus 2 c) EndFraction What is the y-coordinate? StartFraction b c Over 3 EndFraction StartFraction 2 b Over 3 EndFraction StartFraction 2 b c Over 3 EndFraction StartFraction a b c Over 3 EndFraction

Answer 1

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 $\frac{2(a+c)}{3}$. Given that:

$y=\frac{b}{a-2c}x -\frac{2bc}{a-2c} \\\\The\ y\ coordinate\ can\ be \ gotten\ by\ substituting\ the \ value\ of\ x\ and\ simplifying.\\ Substituting\ x:\\\\y=\frac{b}{a-2c}(\frac{2(a+c)}{3} ) -\frac{2bc}{a-2c}$

$Simplyfing\ the\ parenthesis\\y=\frac{2b(a+c)}{3(a-2c)} -\frac{2bc}{a-2c}\\\\y=\frac{2ab+2bc}{3(a-2c)} -\frac{2bc}{a-2c}\\\\Simplyfying\ using\ LCF\\y=\frac{2ab+2bc-6bc}{3(a-2c)}\\\\y=\frac{2ab-4bc}{3(a-2c)}\\\\Factorizing:\\\\y=\frac{2b(a-2c)}{3(a-2c)}\\\\y=\frac{2b}{3}$

The y-coordinate of the intersection point of Line B D and Line C E is at $\frac{2b}{3}$.

Answer 2

Answer:

b

Step-by-step explanation: