Question

prove that ∆XTY ~ ∆YTZ​

Answer 1

Answer:

A=

∣

∣

∣

∣

∣

∣

∣

∣

x+y

z

1

y+z

x

1

z+x

y

1

∣

∣

∣

∣

∣

∣

∣

∣

Applying R

1

→R

1

+R

2

, we get,

A=

∣

∣

∣

∣

∣

∣

∣

∣

x+y+z

z

1

x+y+z

x

1

x+y+z

y

1

∣

∣

∣

∣

∣

∣

∣

∣

A=(x+y+z)

∣

∣

∣

∣

∣

∣

∣

∣

1

z

1

1

x

1

1

y

1

∣

∣

∣

∣

∣

∣

∣

∣

A=(x+y+z)×0=0