Answer:
BP is a diagonal of the square having angle B as one corner
Step-by-step explanation:
You want to show that P(2, 0) lies on the bisector of angle ABC, where A(-4, 2), B(0,-6), and C(6, -3).
Analysis
We can think of perhaps half a dozen ways to show this, none of which seem to be super-simple. The first attachment shows the use of a geometry app to draw the angle bisector of ∠ABC. We notice it passes through point P.
Plotting the points on a graph, we observe that AB has a slope of -2, and BC has a slope of 1/2. Line BP has a slope of 3.
The product of slopes of BA and BC is (-2)(1/2) = -1, which makes angle ABC a right angle. If BP is its bisector, then angles PBA and PBC must both be 45°. The demonstration of this is shown in the third attachment by computing the ratios (vector BA/vector BP) and (vector BP/vector BC). The angle of each vector ratio is 45°, the angle between the vectors in the ratio.
We can use the "distance to a line" formula to show the the distances from P to lines BA and BC are the same. This requires we write equations for those two lines. (Daunting, so we're not ready to do that yet.)
The slope of a line is the tangent of the angle it makes with the x-axis. By finding the angles that BC and BA make with the x-axis, we can compute the angle of the bisector (the average of those two angles) and see if its tangent matches the slope of BP. This is done in the second attachment. The angle computation shows the bisector slope must be 3, which is the slope of line BP.
Angle sum formula
Maybe the simplest method is to use the angle sum formula for tangents to add the angle 45° to the angle of BC. The result should be the slope of BP.
Where α is the angle BC makes with the x-axis, we have ...
tan(α+45°) = (tan(α) +tan(45°))/(1 -tan(α)tan(45°))
= (1/2 +1)/(1 -(1/2)(1)) = (3/2)/(1/2) = 3 . . . . matches the slope of BP
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Additional comment
There are still more ways to show BP is an angle bisector. Point X(4, -4) is a point such that BX and XP are perpendicular and BX=BP. That is BP is a diagonal of the square of which B, X, and P are three corners. Hence BP bisects the right angle at B. (This method simply looks at the graph to observe the slopes and lengths of the various line segments making the square. It requires virtually no math—so takes some discussion to qualify as a "proof".)
We can find point Y where line AC intersects BP, and show that segments AY and CY are proportional to segments BA and BC. This is a condition that is true when BY bisects angle B. (The math for this gets pretty messy.)
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