Find the ratio in which the line \(2x+3y-5=0\) divides the segment joining \(B(8,-9)\) and \(C(2,1)\). Also find the coordinates of the point of division.
Ratio \(8:1\) (internally, from \(B:C\)). Point \(\big(\dfrac{8}{3},\,-\dfrac{1}{9}\big)\).
Step 1: Let the line \(2x+3y-5=0\) divide the segment joining \(A(8,-9)\) and \(B(2,1)\) at \(P(x,y)\) in the ratio \(k:1\) (i.e., \(AP:PB = k:1\)).
Step 2: By the section formula, the coordinates of \(P\) are:
\(x = \dfrac{kx_2 + 1\cdot x_1}{k+1} = \dfrac{k(2) + 8}{k+1} = \dfrac{2k+8}{k+1}\)
\(y = \dfrac{ky_2 + 1\cdot y_1}{k+1} = \dfrac{k(1) + (-9)}{k+1} = \dfrac{k-9}{k+1}\)
Step 3: Since \(P(x,y)\) lies on the line \(2x+3y-5=0\), substitute these values of \(x\) and \(y\):
\(2\left(\dfrac{2k+8}{k+1}\right) + 3\left(\dfrac{k-9}{k+1}\right) - 5 = 0\)
Step 4: Multiply throughout by \((k+1)\):
\(2(2k+8) + 3(k-9) - 5(k+1) = 0\)
Step 5: Expand and simplify:
\(4k + 16 + 3k - 27 - 5k - 5 = 0\)
\((4k+3k-5k) + (16-27-5) = 0\)
\(2k - 16 = 0\)
\(2k = 16 \Rightarrow k = 8\)
Step 6: Hence, the ratio is \(k:1 = 8:1\).
Step 7: Now find the point of division by substituting \(k=8\) in \(P\left(\dfrac{2k+8}{k+1},\dfrac{k-9}{k+1}\right)\):
\(P = \left(\dfrac{2\cdot 8 + 8}{8+1},\dfrac{8-9}{8+1}\right) = \left(\dfrac{24}{9},\dfrac{-1}{9}\right) = \left(\dfrac{8}{3},-\dfrac{1}{9}\right)\)
Final Answer: The line divides the segment internally in the ratio \(8:1\) at \(\left(\dfrac{8}{3},-\dfrac{1}{9}\right)\).