The midpoint of Line AB is M(3, 3). The co-ordinates of B are (4, 7)
Solution:
Given that, the midpoint of Line AB is M(3, 3)
And the coordinates of A are (2, -1)
We have to find what are the coordinates of B
![\text { The midpoint of a line } A\left(x_(1), y_(1)\right) \text { and } B\left(x_(2), y_(2)\right) \text { is given by } M=\left((x_(1)+x_(2))/(2), (y_(2)+y_(2))/(2)\right)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/b4cbrzqz59a3nhsi1w3s6lsiyc3sxzmj1y.png)
![\text { Here } \mathrm{M}=(3,3) \text { and } \mathrm{A}\left(x_(1), y_(1)\right)=(2,-1) \text { and } B\left(x_(2), y_(2)\right)=?](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zrdol1cohxixpiapj83mrdu6tisb79lvp4.png)
Substituting the values we get,
![(3,3)=\left((2+x_(2))/(2), (-1+y_(2))/(2)\right)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5oulp76yinmldxdba8oqk7d7pfmm7kbtqd.png)
![\text { Now, by comparison, } (2+x_(2))/(2)=3 \text { and } (-1+y_(2))/(2)=3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hw71ihahk9j6i8xf161btfpk7bl2c1a5wg.png)
![\begin{array}{l}{\rightarrow 2+x_(2)=6 \text { and }-1+y_(2)=6} \\\\ {\rightarrow x_(2)=6-2 \text { and } y_(2)=6+1} \\\\ {\rightarrow x_(2)=4 \text { and } y_(2)=7}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/24zwnninu7ner0nje2cwsf4qoc29l41zzz.png)
Hence, the co – ordinates of B is (4, 7)