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

b

Explanation:

Answer 2

y = 2b/3

Explanation:

The x-coordinate of the intersection point of Line B D and Line C E is at
`(2(a+c))/(3)`. Given that:

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

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

The y-coordinate of the intersection point of Line B D and Line C E is at
`(2b)/(3)`.