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A line and a circle intersect at  and  as shown below. Find the coordinates of the midpoint of AB.

 

[asy]
usepackage("amsmath");
usepackage("amssymb");

unitsize(2 cm);

pair A, B;

A = dir(240);
B = dir(330);

draw(Circle((0,0),1));
draw(interp(A,B,-0.5)--interp(A,B,1.5));

dot("$A$", A, S);
dot("$B$", B, SE);

label("$x^2 + y^2 = 6$", dir(45), NE);
label("$y = \dfrac{x - 4}{2}$", interp(A,B,1.5), E);
[/asy]



 

 Apr 15, 2024
edited by Hermionegranger  Apr 15, 2024
 #1
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Find the intersection points (A and B):

 

We need to solve the system of equations for x and y:

 

y = (x - 4)/2 (equation of the line)

 

x^2 + y^2 = 6 (equation of the circle)

 

Substitute the first equation for y in the second equation: x^2 + ((x - 4)/2)^2 = 6

 

Solve for x: x^2 + (x^2 - 8x + 16)/4 = 6 4x^2 + x^2 - 8x + 16 = 24 (multiply both sides by 4 to eliminate the fraction) 5x^2 - 8x - 8 = 0 (x - 2)(5x + 4) = 0

 

Therefore, x = 2 or x = -4/5.

 

Find the corresponding y values for each x:

 

For x = 2: y = (2 - 4)/2 = -1

 

For x = -4/5: y = (-4/5 - 4)/2 = -14/5

 

Midpoint coordinates:

 

The midpoint of points A (2, -1) and B (-4/5, -14/5) is: Midpoint = ((2 - 4/5) / 2, (-1 - 14/5) / 2) Midpoint = (6/5, -9/5)

 

Therefore, the midpoint of AB is (6/5, -9/5).

 Apr 15, 2024

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