Question

$$\begin{gathered} Q30Ifx1andy1arerootsof{x}^2+8x-20=0,x2,y2aretherootsof4x^2+32x-57=0 \\ x,yarerootsof9x^2+72x-112=0thenpoints(x1,y1),(x2,y2)and(x3,y3)are \end{gathered}$$

Answer 1

Dear student

$$\begin{gathered} \mathrm{Given}:{\mathrm{x}}_1\mathrm{and}{\mathrm{y}}_1\mathrm{are}\mathrm{roots}\mathrm{of}{\mathrm{x}}^2+8\mathrm{x}-20=0 \\ {\mathrm{x}}_1+{\mathrm{y}}_1=\frac{-\mathrm{coefficient}\mathrm{of}\mathrm{x}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2}=\frac{-8}{1}=-8 \\ {\mathrm{x}}_1{\mathrm{y}}_1=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2}=\frac{-20}{1}=-20 \\ \mathrm{Solving}\mathrm{these}\mathrm{two}\mathrm{equations}\mathrm{we}\mathrm{get}, \\ \left(2,-10\right)\mathrm{or}\left(-10,2\right) \\ \mathrm{Now},\mathrm{Given}:{\mathrm{x}}_2\mathrm{and}{\mathrm{y}}_2\mathrm{are}\mathrm{roots}\mathrm{of}4{\mathrm{x}}^2+32\mathrm{x}-57=0 \\ {\mathrm{x}}_2+{\mathrm{y}}_2=\frac{-\mathrm{coefficient}\mathrm{of}\mathrm{x}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2}=\frac{-32}{4}=-8 \\ {\mathrm{x}}_2{\mathrm{y}}_2=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2}=\frac{-57}{4} \\ \mathrm{Solving}\mathrm{these}\mathrm{two}\mathrm{equations}\mathrm{we}\mathrm{get}, \\ \left(\frac{3}{2},\frac{-19}{2}\right)\mathrm{or}\left(\frac{-19}{2},\frac{3}{2}\right) \\ \mathrm{Now}\mathrm{Given}:{\mathrm{x}}_3\mathrm{and}{\mathrm{y}}_3\mathrm{are}\mathrm{roots}\mathrm{of}9{\mathrm{x}}^2+72\mathrm{x}-112=0 \\ {\mathrm{x}}_3+{\mathrm{y}}_3=\frac{-\mathrm{coefficient}\mathrm{of}\mathrm{x}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2}=\frac{-72}{9}=-8 \\ {\mathrm{x}}_3{\mathrm{y}}_3=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2}=\frac{-112}{9} \\ \mathrm{Solving}\mathrm{these}\mathrm{two}\mathrm{equations}\mathrm{we}\mathrm{get}, \\ \left(\frac{4}{3},\frac{-28}{3}\right)\mathrm{or}\left(\frac{-28}{3},\frac{4}{3}\right) \\ \mathrm{Let}\mathrm{A}=\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(2,-10\right) \\ \mathrm{B}=\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)=\left(\frac{3}{2},\frac{-19}{2}\right) \\ \mathrm{C}=\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)=\left(\frac{4}{3},\frac{-28}{3}\right) \\ \mathrm{Now}\mathrm{AB}=\sqrt{{\left(\frac{3}{2}-2\right)}^2+{\left(\frac{-19}{2}+10\right)}^2}=\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}} \\ \mathrm{BC}=\sqrt{{\left(\frac{4}{3}-\frac{3}{2}\right)}^2+{\left(\frac{-28}{3}+\frac{19}{2}\right)}^2}=\sqrt{\frac{1}{36}+\frac{1}{36}}=\frac{1}{3\sqrt{2}} \\ \mathrm{CA}=\sqrt{{\left(2-\frac{4}{3}\right)}^2+{\left(-10+\frac{28}{3}\right)}^2}=\sqrt{\frac{4}{9}+\frac{4}{9}}=\frac{2\sqrt{2}}{3} \\ \mathrm{Now}\mathrm{AB}+\mathrm{BC}=\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{2}}=\frac{3+1}{3\sqrt{2}}=\frac{4}{3\sqrt{2}}=\frac{2\sqrt{2}}{3}=\mathrm{CA} \\ \mathrm{So}\mathrm{points}\mathrm{are}\mathrm{collinear}. \end{gathered}$$
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