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  • If the coordinates of the mid-points of the sides of a triangle are <img alt="\mathrm{\left ( 1,2 \right )\left ( 0,1 \right )}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%

If the coordinates of the mid-points of the sides of a triangle are \mathrm{\left ( 1,2 \right )\left ( 0,1 \right )} & . Find the coordinates of its vertices. 

Option: 1

\mathrm{A(1,-4), B(3,2) \text { and } C(-1,2)}


Option: 2

\mathrm{A(-1,4), B(2,3) \text { and } C(1,2)}


Option: 3

\mathrm{A(2,-5), B(4,3) \text { and } C(-2,3)}


Option: 4

none of these


Answers (1)

best_answer

Let \mathrm{A\left ( x_{1} , y_{1}\right ), B\left ( x_{2} ,y_{2}\right )} & \mathrm{C\left ( x_{3},y_{3} \right )} be the vertices of \mathrm{\bigtriangleup ABC.} Let \mathrm{D\left ( 1,2 \right ),E\left ( 0,-1 \right ),\ and\ F\left ( 2,-1 \right )} be the mid - points of sides \mathrm{BC, CA \ and\ AB} respectively. Since \mathrm{D} is the mid - point of \mathrm{BC}
\mathrm{\begin{aligned} &\therefore \quad \frac{\mathrm{x}_1+\mathrm{x}_3}{2}=1 \text {, and } \frac{\mathrm{y}_2+\mathrm{y}_3}{2}=2 \\ & \quad \mathrm{y}_3=4 \ldots \text { (i) } \end{aligned} \Rightarrow \mathrm{x}_2+\mathrm{x}_3=2 \text { and } \mathrm{y}_2}
Similarly, E and F are the mid - points of \mathrm{CA} and \mathrm{AB} respectively.

\mathrm{\begin{aligned} & \quad \frac{\mathrm{x}_1+\mathrm{x}_3}{2}=0 \quad \frac{\mathrm{y}_1+\mathrm{y}_3}{2}=-1 \\ & \therefore \quad \text {, and } \quad \ldots \mathrm{x}_1+\mathrm{x}_3=0 \text { and } \mathrm{y}_1 \\ & +\mathrm{y}_3=-2 \quad \ldots \text { (ii) } \\ & \begin{array}{ll} \text { And } \frac{\mathrm{x}_1+\mathrm{x}_3}{2}=2 & \text { and } \frac{\mathrm{y}_1+\mathrm{y}_2}{2}=-1 \\ +\mathrm{y}_2=-2 & \ldots \text { (iii) } \end{array} \mathrm{x}_1+\mathrm{x}_2=4 \text { and } \mathrm{y}_1 \end{aligned}}
Form \mathrm{\left ( i \right )}, \mathrm{\left ( ii \right )} and \mathrm{\left ( iii \right )}, we get \mathrm{\left ( x_{2}+x_{3} \right )+\left ( x_{1}+x_{3} \right )+\left ( x_{1} +x_{2}\right )=2+0+4}
And \quad\left(\mathrm{y}_2+\mathrm{y}_3\right)+\left(\mathrm{y}_1+\mathrm{y}_3\right)+\left(\mathrm{y}_1+\mathrm{y}_2\right)=4-2-2$ \\ $\Rightarrow \quad x_1+x_2+x_3=3$ and $y_1+y_2+y_3=0 .....\mathrm{\left ( iv \right )}
\Rightarrow Form \mathrm{\left ( i \right )} and \mathrm{\left ( iv \right )}, we get \mathrm{x_{1}+2=3} and \mathrm{y_{1}+4=0}
\mathrm{\therefore x_{1}=1} and \mathrm{ y_{1}=-4} So, the coordinates of \mathrm{ A} are \mathrm{ \left ( 1,-4 \right )}
\Rightarrow Form \mathrm{\left (ii \right )} and \mathrm{\left (iv \right )}, we get \mathrm{x_{2}+0=3} and \mathrm{y_{2}-2=0}
\mathrm{\therefore x_{2}=3} and \mathrm{y_{2}=2} So, the coordinates of \mathrm{B} are \mathrm{\left ( 3,2 \right )}
\Rightarrow From \mathrm{\left ( iii \right )} and \mathrm{\left ( iv \right )}, we get \mathrm{x_{3}+4=3} and \mathrm{y_{3}-2=0}
\mathrm{\therefore x_{3}=-1} and \mathrm{y_{3}=2} So, the coordinates of \mathrm{C} are \mathrm{\left ( -1,2 \right )}
Hence, the vertices of the triangle \mathrm{ABC} are \mathrm{A\left ( 1,-4 \right ), B\left ( 3,2 \right )\ and\ C\left ( -1,2 \right ). }

Posted by

vishal kumar

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