In what ratio does the line \(3x-y=5\) divide the segment joining the points \((-4,\,5)\) and \((5,\,-1)\)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2 : 1
Let the line cut the segment in the ratio \(k:1\). The section formula gives the cutting point in terms of \(k\), and that point must satisfy the equation of the line.
Point dividing the segment from \((-4,\,5)\) to \((5,\,-1)\) in the ratio \(k:1\): \(\displaystyle \left(\frac{5k-4}{k+1},\,\frac{-k+5}{k+1}\right)\)
It lies on \(3x-y=5\): \(\displaystyle \frac{3(5k-4)-(-k+5)}{k+1}=5\)
\(\displaystyle \frac{16k-17}{k+1}=5\), so \(16k-17=5k+5\)
\(11k=22\), so \(k=2\)
Ratio \(=2:1\)
Check: the cutting point is \(\displaystyle \left(\frac{10-4}{3},\,\frac{-2+5}{3}\right)=(2,\,1)\), and \(3(2)-1=5\)
