If $P (9a – 2, -b)$ divides the line segment joining $A (3a + 1, -3)$ and $B (8a, 5)$ in the ratio $3 : 1$, find the values of $a$ and $b$.
Given:
$P (9a – 2, -b)$ divides the line segment joining $A (3a + 1, -3)$ and $B (8a, 5)$ in the ratio $3 : 1$.
To do:
We have to find the values of $a$ and $b$.
Solution:
Using the division formula,
$( x,\ y)=( \frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n})$
Here,
$x_1=3a+1,\ y_1=-3,\ x_2=8a,\ y_2=5,\ x=9a-2,\ y=-b, \ m=3$ and $n=1$.
$( 9a-2,\ -b)=( \frac{3\times(8a)+1\times(3a+1)}{3+1},\ \frac{3\times5+1\times(-3)}{3+1})$
$(9a-2,\ -b)=( \frac{24a+3a+1}{4},\ \frac{15-3}{4})$
$( 9a-2,\ -b)=( \frac{27a+1}{4},\ \frac{12}{4})$
This implies,
$9a-2=\frac{27a+1}{4}$ and $-b=3$
$4(9a-2)=27a+1$ and $b=-3$
$36a-8=27a+1$
$36a-27a=8+1$
$9a=9$
$a=1$
Therefore, the values of $a$ and $b$ are $1$ and $-3$ respectively.
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