Question

If point P(9a-2,-b) divides the line segment joining $\text{A (3a+1,-2)}$ and $\text{B }\left( \text{8a,5} \right)$ in the ratio $3:1$. Find the value of a and b.

Answer 1

Hint: To solve this question we should know the concept of section formula. We will find the value of $\text{P}$ coordinates with the help of section formula and then will equate the coordinate value with the value of $\text{P}$given in the question. The section formula is for x and y coordinate is $x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}$and
$y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$ respectively.

Complete step-by-step solution:
The question ask us to find the value of $a$ and $b$ when a line segment having coordinates $\text{A (3a+1,-2)}$ and $\text{B }\left( \text{8a,5} \right)$ and a point $\text{P}$ having coordinate $\text{P }\left( \text{9a-2,-b} \right)$ cuts the line segment in the ratio of $3:1$. Below is the graphical representation of the above line segment.

The part AP is three times that of PB, which means $\text{AP = 3 PB}$. The coordinates of $\text{P}$ can be found with the help of a section formula which states that the coordinate of $\text{P}$is $x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}$ for x- coordinate and for y coordinate the formula is $y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$.
Firstly putting the values in the formulas for x- coordinate. In the question given to us we have $m=3$ and $n=1$ . The coordinates are${{x}_{1}}=3a+1$ and ${{x}_{2}}=8a$. On substituting the values we get:
$x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}$
\[\Rightarrow ~x=\dfrac{3\left( 8a \right)+1\left( 3a+1 \right)}{3+1}\]
$\Rightarrow ~x=\dfrac{24a+3a+1}{4}$
$\Rightarrow ~x=\dfrac{27a+1}{4}$
Now we will find the y coordinates for $\text{P}$ on putting the values in the section formula for it when ${{y}_{1}}=-2$ and ${{y}_{2}}=5$ and values of $m$ and $n$ are same for this coordinate too. On solving this we get:
$y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$
$\Rightarrow ~y=\dfrac{3\left( 5 \right)+1\left( -2 \right)}{3+1}$
$\Rightarrow y=\dfrac{15-2}{4}$
$\Rightarrow y=\dfrac{13}{4}$
So the x-coordinate and y-coordinate of P is $\dfrac{27a+1}{4}$and $\dfrac{13}{4}$ respectively.
In the question the coordinate of $\text{P}$ is given as $\text{9a-2,-b}$ . On equating the coordinates we will get the value of a and b.
On equating the x-coordinate we get:
$\dfrac{27a+1}{4}$= $9a-2$
Taking $4$ to R.H.S we get:
$\Rightarrow 27a+1=4\left( 9a-2 \right)$
$\Rightarrow 27a+1=36a-8$
$\Rightarrow 1+8=36a-27a$
$\Rightarrow 9=9a$
On dividing both sides by $9$ we get:
$\Rightarrow a=1$
Similarly equating the y-coordinate of $\text{P}$ given in the question we get:
$\Rightarrow \dfrac{13}{4}=-b$
On multiplying minus $-1$ both side we get:
$\Rightarrow $ $b=\dfrac{-13}{4}$
$\therefore $ If $\text{P }\left( \text{9a-2,-b} \right)$ divides the line segment joining $\text{A (3a+1,-2)}$ and $\text{B }\left( \text{8a,5} \right)$ in the ratio $3:1$ then value of $a$ and $b$ are $1$ and $\dfrac{-13}{4}$ respectively.

Note: The theorem used to solve the problem is the section formula which is used to determine the coordinate of a point that divides a line segment joining two points into two parts such that the ratio of their length is m:n. internally. The coordinates of $\text{P}$ becomes $x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}$, $y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$when the line segment is divided as given below.