The line joining the points (2, 1) and (5, −8) is trisected at the points

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.

Solution:

We have two points A (2, 1) and B (5,−8). There are two points P and Q which trisect the line segment joining A and B.

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

$\mathrm{P}(x, y)=\left(\frac{n x_{1}+m x_{2}}{m+n}, \frac{n y_{1}+m y_{2}}{m+n}\right)$

The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

$\mathrm{P}\left(x_{1}, y_{1}\right)=\left(\frac{1(5)+2(2)}{1+2}, \frac{2(1)+1(-8)}{1+2}\right)$

$=(3,-2)$

Therefore, co-ordinates of point P is(3,−2)

It is given that point P lies on the line whose equation is

$2 x-y+k=0$

So point A will satisfy this equation.

$2(4)-0+k=0$

So,

$k=-8$

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