Question 1

The midpoint of the line segment from p1 to p2 is (-1,4). If p1 = (-3,6) what is p2? I need help with this please, I am not understanding the reasoning behind the steps taken to solve this. Thank you guys

Question 2

$\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ \square }}\quad ,&{{ \square }})\quad % (c,d) &({{ \square }}\quad ,&{{ \square }}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)$

so.. now, you have the midpoint, and one endpoint, p1 = -3, 6

so... let us use those two fellows

$\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) p1&({{-3 }}\quad ,&{{6}})\quad % (c,d) p2&({{ \square }}\quad ,&{{ \square }}) \end{array}\qquad % coordinates of midpoint \boxed{(-1,4)}\leftarrow midpoint \\\\ \textit{that means, that }\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(-1,4) \\\\ or\ that$

$\bf \begin{cases} \cfrac{{{ x_2}} + {{ x_1}}}{2}=-1\implies &\cfrac{\square +(-3)}{2}=1\\ &\uparrow \\ &\textit{solve for }\square \textit{ to get }x_2\\ \\\\ \cfrac{{{ y_2}} + {{ y_1}}}{2} =4\implies &\cfrac{\square +(6)}{2}=4\\ &\uparrow \\ &\textit{solve for }\square \textit{ to get }y_2\\ \end{cases} \\\\ recall\ that\ p1(x_2,y_2)$

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