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A(– 4, 2), B(6, 4) and C(2, – 2) are the vertices of $$\triangle \mathrm{ABC}$$. Find the equation of right bisector of AB.

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Question

A(– 4, 2), B(6, 4) and C(2, – 2) are the vertices of $$\triangle \mathrm{ABC}$$. Find the equation of right bisector of AB.

Sum
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Solution

Mid-point of $$\mathrm{AB}$$:

$$\mathrm{P} = \left(\frac{-4 + 6}{2}, \frac{2 + 4}{2}\right)$$

$$= (1, 3)$$

Slope of $$\mathrm{AB}$$: 

$$m_{\mathrm{AB}} = \frac{4 - 2}{6 - (-4)}$$

$$= \frac{2}{10}$$

$$= \frac{1}{5}$$ 

Slope of the right bisector of $$\mathrm{AB}$$ is $$-\frac{1}{\frac{1}{5}} = -5$$. 

Passing through $$(1, 3)$$: 

$$y - 3 = -5(x - 1)$$

$$y - 3 = -5x + 5$$

$$5x + y - 8 = 0$$

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Chapter 14: Equation of a Straight Line - EXERCISE 14B [Page 232]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 14 Equation of a Straight Line
EXERCISE 14B | Q 27. (iii) | Page 232
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