\(ABC\) is a triangle, where the vertices \(A\) and \(B\) are fixed points both lying on the \(x\)-axis. Let \(AB = 10\ cm\). The vertex \(C\) moves such that \(\dfrac{1}{\tan A}+\dfrac{1}{\tan B}=\dfrac{1}{k}\), where \(k \ne 0\). What is the equation of the locus of the point \(C\)?
\(y=10k\)
Let \(A=(0,0)\), \(B=(10,0)\) and \(C=(x,y)\), with \(y\) the height from \(C\) to \(AB\). Then \(\tan A=\dfrac{y}{x}\) and \(\tan B=\dfrac{y}{10-x}\), so \(\dfrac{1}{\tan A}+\dfrac{1}{\tan B}=\dfrac{x}{y}+\dfrac{10-x}{y}=\dfrac{10}{y}\). Equating this to \(\dfrac{1}{k}\) gives \(\dfrac{10}{y}=\dfrac{1}{k}\), i.e. \(y=10k\).
Similar Questions
What is the equation of the locus of the mid-point of the line segment obtained by cutting the line x + y = p, (where p is a real number) by the coordinate axes ?
Important Questions from Locus
What is the equation of the locus of the mid-point of the line segment obtained by cutting the line x + y = p, (where p is a real number) by the coordinate axes ?
