Triangle A'B'C' is a reflection of triangle ABC across line BC. Prove that ray BC is the angle bisector of angle ABA'.
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One way to write a line is y=mx+b, where b is a number, m is the slope of the line, and y and x are variables that you can plug numbers into. We know that we have two points, (0,5) and (10,0). To find the slope of a line, we can use the equation
Plugging this in for our points, we get
as our slope (we get -1/2 by dividing both -5 and 10 by 5 from the previous fraction), making our equation y=(-1/2)x+b. Plugging a point in to find out what b is, we get 0=(-1/2)10+b=-5+b. Adding 5 to both sides to separate the b, we get 5=b, making our equation y=(-1/2)x+5. To find out what x is for (x,2), since the y value comes second, we can plug in 2 into our equation to get 2=(-1/2)x+5. Since we want to solve for x, we have to separate it. Subtracting 5 from both sides, we get -3=(-1/2)x. Since we can multiply -1/2 by its reciprocal (switching the numerator and denominator) to get 1 (and therefore x on the right sides as 1*x=x), we multiply both sides by -2 to get 6=x, making the point (6,2)
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Plugging this in for our points, we get
Feel free to ask further questions!