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The line AB has equation 7𝑥 + 2𝑦 = 11

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

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

YEEEEEET  Oct 31, 2018
edited by YEEEEEET  Oct 31, 2018
 #1
avatar+307 
+2

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!

 

Hope this helps! 

Dimitristhym  Oct 31, 2018
edited by Dimitristhym  Oct 31, 2018
 #2
avatar+7515 
+2

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 [\(-1\frac{13}{19}\)\(11\frac{15}{38}\) ]

 

laugh  !

asinus  Oct 31, 2018
edited by asinus  Oct 31, 2018
edited by asinus  Oct 31, 2018
edited by asinus  Oct 31, 2018
edited by asinus  Oct 31, 2018
edited by asinus  Oct 31, 2018

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