If \(P(9a-2,-b)\) divides the segment joining \(A(3a+1,-3)\) and \(B(8a,5)\) in the ratio \(3:1\), find \(a\) and \(b\).
\(a=1,\ b=-3\)
Step 1: Recall the section formula
If a point \(P(x, y)\) divides the line joining two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) in the ratio \(m:n\), then
\[ x = \dfrac{mx_2 + nx_1}{m+n}, \quad y = \dfrac{my_2 + ny_1}{m+n} \]
Step 2: Write the given data
Step 3: Apply section formula for the x-coordinate
\[ x_P = \dfrac{3(8a) + 1(3a+1)}{3+1} = \dfrac{24a + 3a + 1}{4} = \dfrac{27a + 1}{4} \]
But we are also given that \(x_P = 9a - 2\).
So,
\[ 9a - 2 = \dfrac{27a + 1}{4} \]
Step 4: Solve for a
Multiply both sides by 4:
\[ 4(9a - 2) = 27a + 1 \] \[ 36a - 8 = 27a + 1 \] \[ 36a - 27a = 1 + 8 \] \[ 9a = 9 \quad \Rightarrow \quad a = 1 \]
Step 5: Apply section formula for the y-coordinate
\[ y_P = \dfrac{3(5) + 1(-3)}{4} = \dfrac{15 - 3}{4} = \dfrac{12}{4} = 3 \]
But we are given \(y_P = -b\).
\[ -b = 3 \quad \Rightarrow \quad b = -3 \]
Final Answer: \(a = 1, \ b = -3\)