Question:
In the plane with the coordinate system \(Oxy\) Given two lines \({\Delta _1}\) and \({\Delta _2}\) have the equation: \(x – 2y + 1 = respectively: 0\) and \(x – 2y + 4 = 0\), point \(I\left( {2;1} \right).\) Self-centred predicate \(I\) ratio \(k\) turn the line \({\Delta _1}\) into \({\Delta _2}.\) Find \(k.\)
Reference explanation:
Correct Answer: EASY
We take the point \(A\left( {1;1} \right) \in {\Delta _1}.\) Then
\(A’ = {V_{\left( {I,k} \right)}}\left( A \right) \Rightarrow \left\{ {\begin{array}{*{20}{c}}{ x’ = kx + \left( {1 – k} \right)a}\\{y’ = ky + \left( {1 – k} \right)b}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x’ = k + \left( {1 – k} \right)2}\\{y’ = k + \left( { 1 – k} \right)1}\end{array}} \right \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x’ = 2 – k}\\{ y’ = 1}\end{array}} \right.\)
Which \(A’ \in {\Delta _2} \Rightarrow x’ – 2y’ + 4 = 0 \Rightarrow 2 – k – 2.1 + 4 = 0 \Rightarrow k = 4.\)
EASY ANSWER
ADSENSE
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