Question

Find the incenter of the triangle formed by the straight lines y=√3x,y=-√3x,Andy=3

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ coordinates \: of \: incenter \: of \: ΔABC \\ \\ so \: equations \: of \: sides \: AB, \: BC \: ,AC \\ are \: y = \sqrt{3} x \: , \: y = - \sqrt{3} x \: and \: \: y = 3 \\ respectively \\ \\ let \: then \\ y = \sqrt{3} x \: \: \: \: \: \: \: (1)\\ y = - \sqrt{3} x \: \: \: \:(2)\\ y = 3 \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: (3) \\ \\ solving \: (1) \: and \: (2) \\ we \: get \\ A = (x1, \: y1) = (0, \: 0) \\ \\ no w\: solving \: (2) \: and \: (3) \\ we \: get \\ B = (x2, \: y2) = ( \sqrt{3} \:, 3) \\ \\ now \: solving \: (1) \: and \: (3)\\ we \: have \\ C = (x3, \: y3) = ( - \sqrt{3} \:, 3)$

$so \: now \: by \: using \: distance \: formula \\ for \: all \: three \: sides \\ we \: obtain \: that \\ \\ AB = \sqrt{( \sqrt{3} - 0) {}^{2} + (3 - 0) {}^{2} } \\ = \sqrt{3 + 9} \\ = \sqrt{12} \\ \\ BC = \sqrt{(3 - 3) {}^{2} + ( \sqrt{3} - ( - \sqrt{3} )) {}^{2} } \\ = \sqrt{( \sqrt{3} + \sqrt{3} ) {}^{2} } \\ = \sqrt{(2 \sqrt{3} ) {}^{2} } \\ = \sqrt{12} \\ \\ AC = \sqrt{(3 - 0) {}^{2} + ( - \sqrt{3} - 0) {}^{2} } \\ = \sqrt{(3) {}^{2} + ( - \sqrt{3} ) {}^{2} } \\ = \sqrt{9 + 3} \\ = \sqrt{12}$

$thus \: then \: ΔABC \: is \: equilateral \\ in \: nature \\ \\ so \: we \: know \: that \\ for \: an \: equilateral \: triangle \\ incenter \: ,circumcenter \: ,\\ centroid \: ,orthocenter \: ,etc\\ are \: concurrrent \\ \\ thus \: their \: coordinates \: \\ would \: also \: be \: equal$

$so \: we \: know \: that \\ coordinates \: of \: centroid \: are \: \\ calculated \: by \\ x = \frac{x1 + x2 + x3}{3} \: , \: y = \frac{y1 + y2 + y3}{3} \\ \\ = \frac{0 + \sqrt{3} + ( - \sqrt{3} \: )}{3} \: , \: = \frac{3 + 3 + 0}{3} \\ \\ = \frac{ \sqrt{3} - \sqrt{3} }{3} \: , \: = \frac{2 \times 3}{3} \\ \\ = \frac{0}{3} \: , \: = 2 \\ \\ (x,y) = (0,2)$