Question

Given `(ay-bx)/p=(cx-az)/q=(bz-cy)/r`. How to prove that `x/a=y/b=z/c`?

Answer 1

See below.

Explanation:

Calling

`M = ((-b, a, 0),(c, 0, -a),(0, -c, b))`

and

`P = ((x),(y),(z))`

`B=((lambda p),(lambda q),(lambda r))`

we have

`M.P=B`

now we have also

`P_0 = mu ((a),(b),(c))`

If `P = P_0` then

`M.P_0 = B`

`M cdot P_0 = ((0),(0),(0)) = B rArr lambda = 0` so the affirmation

If `M cdot P = B rArr M cdot P_0 = B`

is true when considering `lambda = 0` and `lambda in RR` so it is true

Answer 2

Refer to the Proof given in the Explanation.

Explanation:

Given that,

`(ay-bx)/p=(cx-az)/q=(bz-cy)/r`

`rArr" Each Ratio="{c(ay-bx)+b(cx-az)+a(bz-cy)}/(cp+bq+ar),`

`=0/(cp+bq+ar),`

`=0.`

`rArr (ay-bx)/p=(cx-az)/q=(bz-cy)/r=0.`

`rArr ay=bx, cx=az, bz=cy," or, what is the same as to say, "`

`x/a=y/b, z/c=x/a, y/b=z/c, or, x/a=y/b=z/c.`

Hence, the Proof.

Enjoy Maths.!