question_answer

If the Line segment joining the points \[(3,-4)\] and \[(1,2)\] is triserted at points \[P(a,-2)\] and \[Q\left( \frac{5}{2},b \right),\] then:

A) \[a=\frac{8}{3},\,b=\frac{2}{3}\]          

B) \[a=\frac{7}{3},\,b=0\]

C) \[a=\frac{1}{3},\,b=1\]

D) \[a=\frac{2}{3},\,b=\frac{1}{3}\]

Correct Answer: B

Solution :

[b] Let the line segment joining the point \[A(3,-4)\] and \[B(1,2)\]be trisected by points \[P(a,-2)\] and \[Q\left( \frac{5}{3},b \right)\]

Then,  \[AP=PQ=QB\]

\[\therefore \] P divides AB in the ratio \[1:2\]

So, coordinates of P are

\[\left( \frac{2(3)+1(1)}{1+2},\frac{2(-4)+1(2)}{1+2} \right)\]

i.e., \[\left( \frac{7}{3},-2 \right)\Rightarrow \left( \frac{7}{3},-2 \right)=(a,-2)\]

[\[P=(a,-2)\] (Given)]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,a=\frac{7}{3}\]

Also, Q divides AB in the ratio \[2:1\]

So, coordinates of Q are

\[\left( \frac{1(3)+2(1)}{1+2},\frac{1(-4)+2(2)}{1+2} \right)\]

i.e.,\[\left( \frac{5}{3},0 \right)\]

\[\Rightarrow \,\,\,\,\left( \frac{5}{3},0 \right)=\left( \frac{5}{3},b \right)\]

\[\left[ Q=\left( \frac{5}{3},b \right)(Given) \right]\Rightarrow b=0\]