Find the 13

Find the equation of the circle inscribed in the rhombus ABCD where A[1, -2], B[8, -3] and C[9, 4].

Result

e = (Correct answer is: )

Step-by-step explanation:

$$\begin{gathered} A=(1,-2) \\ B=(8,-3) \\ C=(9,4) \\ S=(A+C)\mathrm{/}2 \\ {x}_0=(A_x+C_x)\mathrm{/}2=(1+9)\mathrm{/}2=5 \\ {y}_0=(A_y+C_y)\mathrm{/}2=((-2)+4)\mathrm{/}2=1 \\ AB: \\ a=1 \\ a\cdot A_x+b\cdot A_y+c=0 \\ a\cdot B_x+b\cdot B_y+c=0 \\ 1\cdot 1+b\cdot (-2)+c=0 \\ 1\cdot 8+b\cdot (-3)+c=0 \\ 2b-c=1 \\ 3b-c=8 \\ b=7 \\ c=13 \\ AB:x+7y+13=0 \\ v=\sqrt{a^2+b^2}=\sqrt{1^2+7^2}=5\sqrt{2}\doteq 7.0711 \\ r={\displaystyle \frac{a\cdot x_0+b\cdot y_0+c}{v}}={\displaystyle \frac{1\cdot 5+7\cdot 1+13}{7.0711}}\doteq 3.5355 \\ R=r^2=3.5355^2={\displaystyle \frac{25}{2}}=12\frac{1}{2}=12.5 \\ e:(x-x_0{)}^2+(y-y_0{)}^2=r^2 \\ e:(x-5{)}^2+(y-1{)}^2=12.5 \end{gathered}$$

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Showing 1 comment:

Ema

I think the coordinate of D is (2,5) and ABCD is a square with side length of 5.sqrt(2) and the circle's radius is 5/sqrt(2). The equation of circle will be (x-5)² + (y-1)² = 25/2

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