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Question: Point P divides the line segment joining the points A(2, 1) and B (5, −8) such that \( \dfrac{AP}{AB...

Point P divides the line segment joining the points A(2, 1) and B (5, −8) such that APAB=13\dfrac{AP}{AB}=\dfrac{1}{3} ​. If P lies on the line 2x – y + k = 0, find the value of k.

Explanation

Solution

Hint: First, find reciprocal of APAB=13\dfrac{AP}{AB}=\dfrac{1}{3} . Then rewrite AB as AP + BP. To get BPAP=2\dfrac{BP}{AP}=2 . Again, take its reciprocal to get APBP=12\dfrac{AP}{BP}=\dfrac{1}{2} . Now, use the section formula to get the coordinates of P=(mc+nam+n,md+nbm+n)P=\left( \dfrac{mc+na}{m+n},\dfrac{md+nb}{m+n} \right) . Substitute the resulting coordinates in 2x – y + k = 0. Solve this to find the value of k which is the final answer.

Complete step-by-step answer:
In this question, we are given that a point P divides the line segment joining the points A (2, 1) and B (5, −8) such that APAB=13\dfrac{AP}{AB}=\dfrac{1}{3} ​.
If P lies on the line 2x – y + k = 0, we need to find the value of k.
We are given that APAB=13\dfrac{AP}{AB}=\dfrac{1}{3}
We will take its reciprocal. Doing this, we will get the following:
ABAP=3\dfrac{AB}{AP}=3
Now, AB can be written as AP + BP. Substituting this in the above expression, we will get the following:
AP+BPAP=3\dfrac{AP+BP}{AP}=3
Now, separating the numerator such that we have two terms in the LHS, we will have the following:
1+BPAP=31+\dfrac{BP}{AP}=3
BPAP=2\dfrac{BP}{AP}=2
Taking the reciprocal of this, we will get the following:
APBP=12\dfrac{AP}{BP}=\dfrac{1}{2}
So, the point P divides the line segment AB in the ratio 1 : 2.
Now we will use the section formula to find the coordinates of the point P.
If point P (x, y) lies on the line segment AB and satisfies AP : PB = m :: n, then we say that P divides AB internally in the ratio m : n. The point of division, P has the coordinates:
P=(mc+nam+n,md+nbm+n)P=\left( \dfrac{mc+na}{m+n},\dfrac{md+nb}{m+n} \right) , where A (a, b) and B (c, d).
So, using the section formula, we will get the following:
P=(5×1+2×22+1,8×1+2×12+1)P=\left( \dfrac{5\times 1+2\times 2}{2+1},\dfrac{-8\times 1+2\times 1}{2+1} \right)
P=(93,63)=(3,2)P=\left( \dfrac{9}{3},\dfrac{-6}{3} \right)=\left( 3,-2 \right)
Now, this point P lies on the line 2x – y + k = 0. Substituting P=(3,2)P=\left( 3,-2 \right) in 2x – y + k = 0, we will get the following:
2×3(2)+k=02\times 3-\left( -2 \right)+k=0
6+2+k=06+2+k=0
k=8k=-8
This is our final answer.

Note: In this question, it is very important to know about the section formula. The section formula tells us the coordinates of the point which divides a given line segment into two parts such that their lengths are in the ratio m : n. If point P (x, y) lies on the line segment AB and satisfies AP : PB = m :: n, then we say that P divides AB internally in the ratio m : n. The point of division, P has the coordinates: P=(mc+nam+n,md+nbm+n)P=\left( \dfrac{mc+na}{m+n},\dfrac{md+nb}{m+n} \right) , where A (a, b) and B (c, d).