Solve Lesson 45 Page 82 SBT Math 10 – Kite>


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.

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

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

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



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