Straight lines

Sumiran agrawal

The image of P(a,b) on y=-x is Q and the image of Q on the line y=x is R . Then what is the midpoint of R?

hemang

let the image of (a,b) on y + x = 0 be (k,m).

so we have (k-a) = (m-b) = -2(a+b)/2.

so, k = 2a + b

and m = 2b + a.

the image of (2a+b,2b+a) on y=x is

(2b+a,2a+b).

but i don't know where this midpoint thing came in between,

how can there be a midpoint of a point??

chaman kirty

i think its about the mid pt of QR....

hemang

then it is easy.

we know two points.

we can easily use the midpoint formula.

Manasi

Let Image of Point P(a,b) wrt y=-x be point Q (h,k)

Then, h=-b and k=-a

i.e. Q(h,k)= Q(-b, -a)

Now, let Image of Point Q(h,k) wrt y=x be point R (m,n)

Then, m=k=-a and n=h=-b

i.e. R(m,n)= R( -a,-b)

Now we have three points, P(a,b) ; Q(-b,-a) ; R(-a, -b)

midpoint of PQ = {(a-b)/2 , -(a-b)/2 } ............which lies on the line y=-x

midpoint of QR = {(-a-b)/2 , (-a-b)/2 } ............which lies on the line y=x

and

midpoint of PR = (0,0) ...which will always be the origin, irrespective of the value of a and b

And one can crosscheck it using geometry also :)