The point A(-6,1) is reflected over the point (-5, 3) and it's image is point B. What are the coordinates of point B?

(a) Distance from A to B:

sqrt[(-6 - (-5))^2 + (1 - 3)^2] = sqrt(1 + 4) = sqrt(5) (Units).


(b) Equation of the straight line through A and B:

(y - 1)/(3 - 1) = (x + 6)/(-5 + 6).

(y - 1)/2 = (x + 6)

y - 1 = 2(x + 6).

y = 2x + 12 + 1.

y = 2x + 13.


(c) Set:

[x - (-5)]^2 + (2x + 13 - 3)^2 = 5.

(x^2 + 10x + 25) + (2x + 10)^2 = 5.

x^2 + 10x + 25 + 4x^2 + 40x + 100 = 5.

5x^2 + 50x + 120 = 0 <--- Divide both sides by 5.

x^2 + 10x + 24 = 0.

(x + 6) * (x + 4) = 0.

Either: x + 6 = 0 ---> x = -6 (we would expect this, as it is the abscissa of point A).

Or: x + 4 = 0 ---> x = - 4. When x = -4, y = 2 * (-4) + 13 = -8 + 13 = 5.


Answer: (-4,5).

Let (h, k) be the coordinates of point B.

The point A(-6, 1) is reflected over the point (-5, 3) and its image is point B(h, k).

(-6 + h)/2 = -5 and (1 + k)/2 = 3
-6 + h = -10 and 1 + k = 6
h = -4 and k = 5

The coordinates of point B = (-4, 5)

A (- 6 ; 1) is reflected over the point M (- 5, 3) and its image is point B (xB ; yB).

So you can deduce that M is the midpoint of [AB]

xM = (xA + xB)/2

2.xM = xA + xB

xB = 2.xM - xA

xB = - 10 + 6

xB = - 4


yM = (yA + yB)/2

2.yM = yA + yB

yB = 2.yM - yA

yB = 6 - 1

yB = 5

if you plot it out on graph paper, you will better see what is happening

B (- 4 , 5 )

(not sure)
(-4,-1)
https://gyazo.com/840b97968ed40ca221943b3ca40617ff