Question

The. Line. Segment. Joining. The points. A. (2,1). And. B. (5,-8). Is. Trisected. At. The. Points. P. And. Q. Such. That. P. Is. Nearer. To. A if. P. Also. Line. On. The. Line. Given. By. 2x-y+0. Find. The. Value. Of. K

Answer 1

Given: Points are A(2,1) and B (5,-8), P and Q trisect the line segment AB,  equation is 2x-y+k=0

To find: The value of k?

Solution:

• Now, as we have given that P and Q trisects the line AB, so we can conclude that:

              AP = PQ = QB

• Now, PB = PQ + QB

• Which can also be written as PB = AP + AP = 2AP as all the three are equal.

• So, ratio of:

             AP / PB = AP / 2AP

             1/2 = 1:2

• Now, P divides AB in ratio 1:2, so by ratio formula, we get:

            (mx2+nx1 / m+n)  , (my2+ny1  /m+n)

• Putting values, we get:

            1(5) + 2(2) / 1+2 , 1(-8) + 2(1)/ 1+2

            (5+4)/3 , -8+2/3)

            (3,-2)

• As of now, we have got the coordinates of P so putting them in the equation, we get:

            2(3) - (-2) + k = 0

            6 + 2 + k = 0

            k = -8

Answer:

                So, the value of k is -8.