Question

Please solve the question in the attachment:--​

Answer 1

Question:

If px + qy + r = 0 and qx + py + r = 0 (x≠y) , then show that the value of (x + y) is (-r/p) or (-r/q) .

Given:

px + qy + r = 0

qx + py + r = 0

To prove:

x + y = -r/p or -r/q

Proof:

Let ,

px + qy + r = 0 ---------(1)

qx + py + r = 0 ------------(2)

Now,

Subtracting eq-(2) from eq-(1) ,we get;

=> (px + qy + r) - (qx + py + r) = 0

=> px + qy + r - qx - py - r = 0

=> px - qx + qy - py = 0

=> x(p - q) - y(p - q) = 0

=> (p - q)(x - y) = 0

Either (p - q) = 0 or (x - y) = 0

ie; (p = q) or (x = y)

But,

It is given that, x ≠ y , thus the case

x = y will be rejected.

Thus,

We have , p = q .

Now,

Adding eq-(1) and eq-(2) ,we get;

=> (px + qy + r) + (qx + py + r) = 0

=> px + qy + r + qx + py + r = 0

=> px + qx + qy + py + 2r = 0

=> x(p + q) + y(p + q) + 2r = 0

=> (p + q)(x + y) + 2r = 0

=> (p + q)(x + y) = - 2r

=> x + y = - 2r/(p+q)

=> x + y = - 2r/(p+p) { p = q }

=> x + y = - 2r/2p

=> x + y = - r/p

Also,

=> x + y = -r/q { p = q }

Hence,

x + y = - r/p or - r/q .

Answer 2

Answer:

Step-by-step explanation: