Question

point T is in the interior of rectangle PQRS prove that TS + TQ = TP + TR

Answer 1

Answer:

(TS+TQ)= (TP+TR).... proved

Step-by-step explanation:

Let us assume that any rectangle PQRS(a×b) is placed on the coordinate plane in such a way that the coordinates of P(0,0), Q(a,0), R(a,b) and S(0,b) respectively. (See the diagram)

Also assume an interior point T(h,k) to the rectangle on the coordinate plane.

Then, we can have

$TS=\sqrt{h^{2}+(k-b)^{2}}$ ...... (1)

$TQ=\sqrt{(h-a)^{2}+k^{2}}$ ...... (2)

$TP=\sqrt{h^{2}+k^{2}}$ ...... (3)

and, $TR=\sqrt{(h-a)^{2}+(k-b)^{2}}$ ..... (4)

We have to prove that, TS+TQ= TP+TR

Now, $(TS+TQ)^{2}=(\sqrt{h^{2}+(k-b)^{2}}+\sqrt{(h-a)^{2}+k^{2}})^{2}$

⇒$(TS+TQ)^{2}= h^{2}+k^{2}+(h-a)^{2}+(k-b)^{2}+2\sqrt{(h^{2}+(k-b)^{2})((h-a)^{2}+k^{2}  )  }$

⇒$(TS+TQ)^{2} =h^{2}+k^{2}+(h-a)^{2}+(k-b)^{2}+2\sqrt{h^{2}(h-a^{2})+h^{2}k^{2}+k^{2}(k-b)^{2}+(h-a)^{2}(k-b)^{2}}$ .... (5)

Similarly, $(TP+TR)^{2}=(\sqrt{h^{2}+k^{2}}+\sqrt{(h-a)^{2}+(k-b)^{2}})^{2}$

⇒$(TP+TR)^{2}=h^{2}+k^{2}+(h-a)^{2}+(k-b)^{2}+2\sqrt{h^{2}(h-a^{2})+h^{2}k^{2}+k^{2}(k-b)^{2}+(h-a)^{2}(k-b)^{2}}$ .... (6)

So, from (5) and (6) it is clear that,

$(TS+TQ)^{2}=(TP+TR)^{2}$

⇒(TS+TQ)= ±(TP+TR)

⇒(TS+TQ)= (TP+TR) {Since sum of two lengths can not be negative}

Hence, proved.