The line AB has equation 7𝑥 + 2𝑦 = 11

Find the distance between C and midpoint of AC with the \(\sqrt{(χ2-χ1)^2+(y2-y1)^2}\) And with this resault is equal with \(\sqrt{(χ3-χ2)^2+(y3-y2)^2}\) (the distance between midpoint of AC and A) and you have x3,y3 unknowns and you have and the equation 7𝑥 + 2𝑦 = 11 so 2 equations and 2 unknowns.So you solve it! Try it alone!

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The line AB has equation 7𝑥 + 2𝑦 = 11

The point ( 5/ 2 , 10) (x1 , y1) is the midpoint of AC. Find the coordinates of point A

The point C has coordinates (-5, 25 /2 ) (x2 , y2)

Hello YEEEEEET !

Two-point form

\(y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1\\ y=\frac{12.5-10}{-5-2.5}(x-2.5)+10\\ y=-\frac{1}{3}(x-2.5)+10\\ y=-\frac{1}{3}x+10\frac{5}{6}\\ \color{blue}y=-\frac{1}{3}x+\frac{65}{6}\)

\(7x+2y=11\\ y=-3.5x+5.5\\ \color{blue}y=-\frac{7}{2}x+\frac{11}{2}\)

\(-\frac{1}{3}x+\frac{65}{6}=-\frac{7}{2}x+\frac{11}{2}\\ (-\frac{1}{3}+\frac{7}{2})x=\frac{11}{2}-\frac{65}{6}\\ \frac{19}{6}x=-\frac{32}{6}\\ 19x=-32\\ x=-\frac{32}{19}\\ \color{blue}x=-1\frac{13}{19}\)

\(y=-\frac{7}{2}x+\frac{11}{2}\\ y=\frac{-7x+11}{2}\\ y=\frac{-7\cdot (-\frac{32}{19})+11}{2}\\ y=\frac{\frac{224}{19}+11}{2}\\ y=\frac{433}{38}\\ \color{blue}y=11\frac{15}{38}\)

The coordinates of point A are [\(-1\frac{13}{19}\), \(11\frac{15}{38}\) ]

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