Question

The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q. If point P lies on the line 2x – y + k = 0. Find the value of k.

Answer 1

Hint: Assume the given points as A (2, 1) and B (5, -8). Plot the points P and Q on the line joining A, B such that AP = PQ = QB. Now, apply the section formula given as: - \[x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}\] and \[y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\] to determine the coordinates of P. Here, (x, y) is the assumed coordinates of P, m : n = 1 : 2 is the ratio in which P divides AB and \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] are the coordinates of A and B respectively. Now, substitute the obtained value of x and y in the equation of the line 2x – y + k = 0 and solve for the value of k to get the answer.

Complete step-by-step answer:
Here, we have been provided with two points (2, 1) and (5, 8) and it is said that the line connecting these points is trisected by the points P and Q.
Now, let us assume these points as A (2, 1) and B (5, 8). So, according to the given situation, we have,

Here, P and Q are trisecting the line joining A, B. That means AB is divided into three equal parts.
\[\Rightarrow \] AP = PQ = QB – (1)
\[\Rightarrow \dfrac{AP}{PB}=\dfrac{AP}{PQ+QB}\]
Using relation (1), we have,
\[\Rightarrow \dfrac{AP}{PB}=\dfrac{AP}{AP+AP}=\dfrac{AP}{2AP}=\dfrac{1}{2}\]
So, we can say that point P is dividing the line AB in the ratio 1 : 2. Let us assume the coordinates of P as (x, y). Now, we know that the section formula states that if a point (x, y) divides a line segment joining two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] in the ratio m : n then the value of x and y is given as: -
\[\Rightarrow x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}\]
\[\Rightarrow y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\]
So, in the above figure, we have,
\[\Rightarrow \] P = (x, y)
\[\Rightarrow \] A = (2, 1) = \[\left( {{x}_{1}},{{y}_{1}} \right)\]
\[\Rightarrow \] B = (5, -8) = \[\left( {{x}_{2}},{{y}_{2}} \right)\]
\[\Rightarrow \] m : n = 1 : 2
Therefore, applying section formula, we get,
\[\begin{align}
  & \Rightarrow x=\dfrac{1\times 5+2\times 2}{1+2} \\
 & \Rightarrow x=\dfrac{5+4}{3} \\
 & \Rightarrow x=3 \\
 & \Rightarrow y=\dfrac{1\times \left( -8 \right)+2\times 1}{1+2} \\
 & \Rightarrow y=\dfrac{-8+2}{3} \\
 & \Rightarrow y=-2 \\
\end{align}\]
Hence, the coordinates of P is given as P (3, -2).
Now, it is given that point P lies on the line 2x – y + k = 0, so it must satisfy the equation of the given line. Therefore, substituting P (3, -2) in the given equation of line, we get,
\[\begin{align}
  & \Rightarrow 2\times 3-\left( -2 \right)+k=0 \\
 & \Rightarrow 6+2+k=0 \\
 & \Rightarrow k=-8 \\
\end{align}\]
Hence, the value of k is -8.

Note: One may note that we can also determine the coordinates of point Q by assuming the ratio as 2 : 1. But the coordinates of point Q are of no use here because it will not affect the coordinates of P. Remember the section formula to solve the above question. You must consider m and n carefully otherwise you will get the wrong answer.