Question

If A (1, -1), B( -4, 6)and C (1, 6) are the vertices of ΔABC, then the coordinatesof its circumcentre is

Answer 1

Answer:

your solution is as follows

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refer the reference diagram uploaded above

Step-by-step explanation:

$to \: find : \\ coordinates \: of \: circumcentre \\ of \: ΔABC \\ \\ so \: let \: here \: O \: be \: the \: circumcentre \\ of \: ΔABC \\ such \: that \\ O = (x,y) \\ \\ so \: let \: here \\ slope \: of \: AB \:b e \: m1 \: and \\ that \: of \: OP \: be \: m2 \\ \\ so \: we \: have \\ slope \: of \: AB = \frac{y2 - y1}{x2 - x1} \\ \\ = \frac{6 - ( - 1)}{ - 4 - 1} \\ \\ = \frac{6 + 1}{ - 5} \\ \\ = \frac{ - 7}{5} \\ \\ but \: here \: \\ ∵AB \: ⊥ \: OP \\ \\ m1 \times m2 = - 1 \\ \\ \frac{ - 7}{5} \times m2 = - 1 \\ \\ m2 = \frac{5}{7}$

$moreover \: since \: \\ P \: is \: midpoint \: of \: AB \\ \\ by \: midpoint \: formula \\ we \: have \\ \\ P = ( \frac{x1 + x2}{2} \: , \: \frac{y1 + y2}{2} \: ) \\ \\ = \frac{1 - 4}{2} \: , \: \frac{6 - 1}{2} \\ \\ P = ( \frac{ - 3}{2} \: ,\: \frac{5}{2} \: )$

$now \: we \: have \\ slope \: intercept \: equation \: of \: \\ any \: straight \: line \\ is \: given \: by \\ \\ y - y1 = m(x - x1) \\ \\ so \: then \\ equation \: of \: OP \: is \\ \\ y - \frac{3}{2} = \frac{5}{7} (x + \frac{3}{2} ) \\ \\ \frac{2y - 3}{2} = \frac{5}{7} ( \frac{2x + 3}{2} ) \\ \\ 7(2y - 5) =5( 2x + 3) \\ \\ 14y - 35 = 10x + 15 \\ \\ 10x - 14y = - 50 \\ \\ 5x - 7y = - 25 \: \: \: \: \: \: \: \: \: \: (1)$

$similarly \: then \: \: let \\ \: slope \: of \: AC \: be \: a \: and \: that \: of \\ OR\: \: be \: b \\ \\ slope \: of \: AC = \frac{y2 - y1}{x2 - x1} \\ \\ = \frac{6 - ( - 1)}{1 - 1} \\ \\ = \frac{6 + 1}{0} \\ \\ a= \frac{7}{0}$

$now \: since \\ AC \: ⊥ \: OR \\ \\ a \times b = - 1 \\ \\ \frac{7}{0} \times b = - 1 \\ \\ 7b = 0 \\ \\ b = 0$

$so \: accordingly \\ as \: here \\ R \: is \: midpoint \: of \: AC \\ \\ R = ( \frac{y1 + y2}{2} ) \\ \\ = \frac{6 - 1}{2} \\ \\ R = y1= \frac{5}{2} \\ \\ so \: we \: have \\ y - y1 = m(x - x1) \\ \\ y - \frac{5}{2} = 0(x - x1) \\ \\ y - \frac{5}{2} = 0 \\ \\ y = \frac{5}{2} \: \: \: \: \: \: \:$

$substituting \: value \: of \: \: y \: in \: (1) \\ we \: get \\ \\ x = \frac{ -3}{ 2} \\ \\ henceforth \: here \\ \\ O = ( \frac{ - 3}{2} \: , \: \frac{5}{2} \: )$