Question

If $\frac{x}{q+r-p}=\frac{y}{r+p-q}=\frac{z}{p+q-r},$ shew that (q - r)x + (r - p)y + (p - q) z = 0.

Answer 1

To Show that

$(q-r)x+(r-p)y+(p-q)z=0$

Here , we have

$\frac{x}{q+r-p}=\frac{y}{r+p-q}=\frac{z}{p+q-r}$

Let ,

$\frac{x}{q+r-p}=k$

$\frac{y}{r+p-q}=k$

$\frac{z}{p+q-r}=k$

So that ,

$x=k(q+r-p),y=k(r+p-q),z=k(p+q-r)------>Equation1$

Taking LHS

$=(q-r)x+(r-p)y+(p-q)z$

Putting the value of x , y and z From Equation 1 , we get

$=(q-r)\times k(q+r-p)+(r-p)\times k(r+p-q)+(p-q)\times k(p+q-r)$

$=k[q^2-r^2-pq+pr]+k[r^2-p^2-qr+pq]+k[p^2-q^2-pr+rq]$

$=k[q^2-q^2+p^2-p^2-pr+pr-qr+qr-pq+pq]$

$=k\left[0\right]$

$=0$

$=$ RHS