Question

Please answer this question with steps . The one who answers correclty with clear explanation will be marked as brainliest. ( irrelevant answers will be reported ) ​

Answer 1

Step-by-step explanation:

Given, the point P[{x1 + t(x2 - x1 )}, {y1 + t (y2 - y1 )}] divides the join of A(x1 ,y1 ) and B(x2 ,y2 ) internally.

Let the ratio is k : 1

Now, {x1 + t(x2 - x1 ), y1 + t (y2 - y1 )} = {(kx2 + x1 )/(k + 1), (ky2 + y1 )/(k + 1)}

Now, x1 + t(x2 - x1 ) = (kx2 + x1 )/(k + 1)

=> x1 (1 - t) + tx2 = kx2 /(k + 1) + x1 /(k + 1)

Now, equate coefficient of x2 , we get

=> t = k/(k + 1)

Since k > 0 {since ratio can not be negative}

=> 0 < k/(k + 1) < 1

=> 0 < t < 1

Hence, the value of t always lies between 0 and 1

Answer 2

Answer:

P[[x1 + t{x2 - x1 )), {y1 +t (V2 - y1)}} divides the join of A(x1 y1) and B(x2 y2 ) internally.

ratio is k :1

Now, (x1+ tx2-x1), y1+y (y2-y)) = {(kx2 + x1 /(k + 1). (ky2+y1y/(k + 1)]

Now, x1 + t(x2 - x1) = (kx2 + x1 /(k + 1)

=> x1 (1 t) + tx2 = kx2 /(k + 1) + x 1 /(k + 1)

Now, equate coefficient of x2, we get

=> t = k/(k + 1)

Since k>0..

=> 0 < k/(k + 1) <1

=>0<t<1

So..it's proved that value of t always lies between 0 and 1...

Hope it helps you...