Question

The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q . If the coordinates of P and Q are (p, -2) and (5/3, q) respectively. Find the values of p and q.

Answer 1

The line segment joining the points (3,-4) and (1,2) is trisected  at the points P and Q.

• Coordinates of P : (p, -2)
• Coordinates of Q : (5/3, q)

From the section formula for P,

P(x,y) = ( ( 1*1 + 2*3 )/1+2  ,  ( 1*2 + -4*2)/1+2 ) = (7/3, -2)

Q(x,y) = ( ( 2*1 + 1*3 )/1+2  ,  ( 2*2 + -4*1)/1+2 ) = (5/3, 0)

Hence, p=7/3 and q=0.  

Answer 2

The line segment joining the points (3,-4) and (1, 2) is trisect at a point P(p, - 2) and Q (5/3,q)

The value of p=7/3 and the value of q =0

Step 1:

Let's consider the joining point of the straight line is A(3,-4) and B(1,2)

Step 2:

As at point P and Q the straight line AB is trisect so for straight line AQ P is the mid point.

Step 3:

As P is the mid point of AQ so

X co ordinate of P = ( X co ordinate of A+ X co ordinate of Q) /2

Similarly

Y co ordinate of P = ( Y co ordinate of A+ Y co ordinate of Q) /2.

Step 4:

For calculating X co ordinate

p = (3+5/3)/2

p= 7/3

For calculating Y co ordinate

-2=(-4+q)/2

q=0

Therefore the value of p=7/3

and the value of q=0

Step 5:

To cross check the answer if we consider the PB straight line and Q is the mid point of PB. Then do the the problem in similar method. The value of p will come 7/3 and value of q will come 0.