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Brief Review of Cartesian System of Rectangular Coordinates — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSBrief Review of Cartesian System of Rectangular Coordinates1 Mark
Question
If the points (1, -1), (2, -1) and (4, -3) are the mid-points of the sides of a triangle, then write the coordinates of its centroid.
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Answer

Let $\text{p}(1_1-1),\ \text{Q}(2_1-1)\ \text{R}(4_1-3)$ be the mid-points of sides AB, BC and CA respectively of triangle ABC. Let $\text{A}(\text{x}_1,\ \text{y}_1),\ \text{B}(\text{x}_2,\ \text{y}_2)$ and $\text{C}(\text{x}_3,\ \text{y}_3)$ be the vertices P is the mid-point of AB $\therefore\frac{\text{x}_1+\text{x}_2}{2}=1,\ \frac{\text{y}+\text{y}_2}{2}=-1$ $\Rightarrow\text{x}_1+\text{x}_2=2$ and $\text{y}_1+\text{y}_2=-2\ ......(1)$ Q is the mid-point of BC $\therefore\frac{\text{x}_2+\text{x}_3}{2}=2,\ \frac{\text{y}_2+\text{y}_3}{2}=-1$ $\Rightarrow\text{x}_2+\text{x}_3=4$ and $\text{y}_2+\text{y}_3=-2\ ....... (2)$ R is the mid-point AC $\therefore\frac{\text{x}_1+\text{x}_3}{2}=4,\ \frac{\text{y}_1+\text{y}_3}{2}=-3$ $\Rightarrow​​\text{x}_1+\text{x}_2=8$ and $\text{y}_1+\text{y}_2+\text{y}_3=-6\ .......(3)$ Adding equations (i), (ii) and (iii), we get $\text{x}_1+\text{x}_2+\text{x}_2+\text{x}_3+\text{x}_1+\text{x}_3=2+4+8$ and, $\text{y}_1+\text{y}_2+\text{y}_2+\text{y}_3+\text{y}_1+\text{y}_3=-2-2-6$ $\Rightarrow2(\text{x}_1+\text{x}_2+\text{x}_3)=14$ and, $2(\text{y}_1+\text{y}_2+\text{y}_3)=-10$ $\Rightarrow\text{x}_1+\text{x}_2+\text{x}_3=7$ and $\text{y}_1+\text{y}_2+\text{y}_3=-5$ $\therefore$ Coordinates of centroid are $\Big(\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3},\ \frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)$ i.e $\Big(\frac{7}{3},\ \frac{-5}{3}\Big).$

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