Question

The line segment joining the point 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 lies on the line given by 2x – y + k = 0, find the value of k.

Answer 1

Let the points A(2, 1) and B(5, –8) is trisected at the points P(x, y) and Q(a, b).

Thus, AP = PQ = QB

Therefore, P divides AB internally in the ratio 1 : 2.

Section formula: if the point (x, y) divides the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio m : n, then the coordinates (x, y) = $\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

Therefore, using section formula, the coordinates of P are:

$$\begin{gathered} \left(x,y\right)=\left(\frac{1\left(5\right)+2\left(2\right)}{1+2},\frac{1\left(-8\right)+2\left(1\right)}{1+2}\right) \\ \Rightarrow \left(x,y\right)=\left(\frac{5+4}{3},\frac{-8+2}{3}\right) \\ \Rightarrow \left(x,y\right)=\left(\frac{9}{3},\frac{-6}{3}\right) \\ \Rightarrow \left(x,y\right)=\left(3,-2\right) \end{gathered}$$

Hence, the coordinates of P are (3, –2).

Since, P also lies on the line given by 2x – y + k = 0,
Therefore, (3, –2) satisfies the equation 2x – y + k = 0
$$\begin{gathered} 2\left(3\right)-\left(-2\right)+k=0 \\ \Rightarrow 6+2+k=0 \\ \Rightarrow k=-8 \end{gathered}$$

Hence, the values of k is –8.