**Topic**

Give three points *A*(-2; 2), *REMOVE*(4 ; 2), *OLD*(sixty four). Write the equation of the line passing through *REMOVE* at the same time equidistant *A* and *OLD*.

**Solution method – See details**

Let the line ∆ have the form: *ax* + *by* + *c* = 0 (1)

Step 1: Change the coordinates *REMOVE* enter PT (1) and perform *c* according to the *a* and *b*

Step 2: Use the distance formula to create a PT of the form *d*(*Ah,*) = *d*(*C,*)

Step 3: Solve the above PT to find the relationship between *a* and *b*

Step 4: Select 2 values *a* and *b* follow the relationship then write PT

**Detailed explanation**

Suppose ∆ has the form: *ax* + *by* + *c* = 0 (1)

Since \(B(4;2) \in \Delta \) so \(4a + 2b + c = 0 \Rightarrow c = – 4a – 2b\)\( \Rightarrow \Delta :ax + by – 4a – 2b = 0\)

According to the assumption, *d*(*Ah,*) = *d*(*C,*) \( \Leftrightarrow \frac{{\left| { – 2a + 2b – 4a – 2b} \right|}}{{\sqrt {{a^2} + {b^2}} }} = \frac{ {\left| {6a + 4b – 4a – 2b} \right|}}{{\sqrt {{a^2} + {b^2}} }}\)

\( \Rightarrow \left| { – 6a} \right| = \left| {2a + 2b} \right| \Leftrightarrow 6\left| a \right| = \left| {2a + 2b} \right| \Leftrightarrow \left[\begin{array}{l}6a=2a+2b\\6a=–2a–2b\end{array}\right\)[\begin{array}{l}6a=2a+2b\\6a= –2a–2b\end{array}\right\)

\( \Leftrightarrow \left[\begin{array}{l}4a=2b\\8a=–2b\end{array}\right\Leftrightarrow\left[\begin{array}{l}2a=b\\–4a=b\end{array}\right\)[\begin{array}{l}4a=2b\\8a= –2b\end{array}\right\Leftrightarrow\left[\begin{array}{l}2a=b\\–4a=b\end{array}\right\)

+ With 2*a* = *b*select \(a = 1 \Rightarrow b = 2\)\( \Rightarrow \) ∆ with PT: *x* + 2*y* – 8 = 0

+ With -4*a* = *b*select \(a = 1 \Rightarrow b = – 4\)\( \Rightarrow \) ∆ with PT: *x* – 4*y* + 4 = 0

So there are 2 lines satisfying *x* + 2*y* – 8 = 0 and *x* – 4*y* + 4 = 0

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