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Question Number 80207 by peter frank last updated on 01/Feb/20

Commented by john santu last updated on 01/Feb/20

$$\left(2\right)area=\frac{1}{2}\mid \left|\begin{array}{c}21\\ 3-2\end{array}\right|+\left|\begin{array}{c}3-2\\ xx+3\end{array}\right|+\left|\begin{array}{c}xx+3\\ 21\end{array}\right|\mid$$$$\Rightarrow 5=\frac{1}{2}\mid (-7)+(5x+9)+(-x-6)\mid$$$$\Rightarrow 10=\mid 4x-4\mid$$$$\pm 5=2x-2,2x=2\pm 5\Rightarrow \{\begin{array}{l}x=\frac{7}{3},y=\frac{16}{3}\\ x=-\frac{3}{2},y=\frac{3}{2}\end{array}$$$$thereforethethirdverticestriangle$$$$is(\frac{7}{3},\frac{16}{3})and(-\frac{3}{2},\frac{3}{2})$$$$$$$$$$

Commented by mr W last updated on 01/Feb/20

$$question1)iswrong.thepoints$$$$shouldbe(a,-a),(-a,a),(\sqrt{3}a,\sqrt{3}a).$$$$justdrawthepointsyouwillsee.$$

Commented by peter frank last updated on 01/Feb/20

$$thankyousircollection$$$$thenhowtoshowifitis$$$$equalateral?$$

Commented by mr W last updated on 01/Feb/20

$$sayA(a,-a),B(-a,a),C(\sqrt{3}a,\sqrt{3}a)$$$$AB=\sqrt{{\left(2a\right)}^2+{\left(2a\right)}^2}=2\sqrt{2}a$$$$AC=\sqrt{{(\sqrt{3}-1)}^2{a}^2+{(\sqrt{3}+1)}^2{a}^2}=2\sqrt{2}a$$$$BC=\sqrt{{(\sqrt{3}+1)}^2{a}^2+{(\sqrt{3}-1)}^2{a}^2}=2\sqrt{2}a$$$$\Rightarrow AB=AC=BC$$$$\Rightarrow equilateral!$$

Commented by kaivan.ahmadi last updated on 01/Feb/20

$$A=(a,a),B=(-a,a),C=(3a,3a)$$$$leta>0$$$$\mid AB\mid =\sqrt{({(a+a)}^2}=2a$$$$\mid AC\mid =\sqrt{({\left(2a\right)}^2+{\left(2a\right)}^2}=2\sqrt{2}a$$$$\mid BC\mid =\sqrt{{\left(4a\right)}^2+{\left(2a\right)}^2}=\sqrt{20a^2}=2\sqrt{5}a$$$$so△ABCisnotequalateraltriangle$$$$$$

Commented by peter frank last updated on 01/Feb/20

$$Iappriciateyourtimesir.thankyou$$$$mrw\mathrm{\&}kaivan$$

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