Question

If A ( -2 ,1 ), B ( 2, 3 ) and C ( 1, 1 ), find the equation of the line which bisects the angle B (that is ∠).​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ equation \: of \: line \: bisecting \: ∠ABC \\ \\ so \: let \: the \: line \: which \: bisects \\ ∠ABC \: \: be \: \: BP \\ \\ so \: since \\ \: BP \: bisects \: ∠ABC \\ \\ by \: angle \: bisector \: property \\ we \: have \\ \\ \frac{AB}{BC} = \frac{AP }{ CP} \: \: \: \: \: \: \: (1)$

$so \: we \: know \: that \\ distance \: formul a \: is \: given \: by \\ \\ \sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} } \\ \\ so \: thus \: then \\ \\ d(AB) = \sqrt{(2 - ( - 2)) {}^{2} + (3 - 1) {}^{2} } \\ \\ = \sqrt{(2 + 2) {}^{2} + (2) {}^{2} } \\ \\ = \sqrt{(4) {}^{2} + (2) {}^{2} } \\ \\ = \sqrt{16 + 4 } \\ \\ = \sqrt{20} \\ \\ AB = 2 \sqrt{5} . \\ \\ similarly \\ \\ d(BC) = \sqrt{(1 - 2) {}^{2} + (1 - 3) {}^{2} } \\ \\ = \sqrt{( - 1) {}^{2} + ( - 2) {}^{2} } \\ \\ = \sqrt{1 + 4} \\ \\BC = \sqrt{5}$

$now \: dividing \: AB \: \: and \: \: BC \\ we \: get \\ \\ \frac{AB}{BC} = \frac{2 \sqrt{5} }{ \sqrt{5} } \\ \\ \frac{AB}{BC} = \frac{2}{1} \\ \\ so \: thus \: then \\ \\ \frac{AP }{ CP} = \frac{2}{1} \: \: \: \: \: \: \: \: \: from \: (1)$

$so \: thus \\ P \: \: divides \: \: seg \:AC \: internally \\ in \: ratio \: \: 2 : 1 \\ \\ so \: let \: then \\ AP : CP = m : n = 2 : 1$

$we \: know \: that \\ section \: formula \: for \: internal \\ division \: \: is \: given \: by \\ \\ x = \frac{mx2 + nx1}{m + n} \: \: ; \: \: y = \frac{my2 + ny1}{m + n} \\ \\ so \: hence \\ \\ P = \frac{(2 \times 1) + 1 \times ( - 2)}{(2 + 1)} \: \: ; = \frac{(2 \times 1) + (1 \times 1)}{2 + 1} \\ \\ = \frac{2 - 2}{3} \: \: ; = \frac{2 + 1}{3} \\ \\ = \frac{0}{3} \: \: ; = \frac{3}{3} \\ \\ P = (x,y) = (0,1)$

$so \: thus \: then \\ let \: here \\ (x1,y1) = (2,3) \\ (x2,y2) = (0,1) \\ \\ we \: know \: that \\ two \: points \: equation \: of \: straight \\ line \: is \: given \: by \\ \\ \frac{x - x1}{x1 - x2} = \frac{y - y1}{y1 - y2} \\ \\ \frac{x - 2}{2 - 0} = \frac{y - 3}{3 - 1} \\ \\ \frac{x - 2}{2} = \frac{y - 3}{2} \\ \\ x - 2 = y - 3 \\ \\ x - y + 1 = 0 \\ \\ or \\ \\ x - y = - 1$

$hence \: the \: \: equation \: of \: line \: \\ bisecting \: \: ∠ABC \: \: is \\ x - y + 1 = 0$