# Technical Document Extraction: Mathematical Problem and Solution ## 1. Document Overview This image contains a mathematical word problem followed by a step-by-step transcript of a solution process. The text includes mathematical notation in LaTeX format and "score" markers, suggesting it is a transcript from an automated reasoning or tutoring system. --- ## 2. Component Isolation ### Region 1: Header (Problem Statement) **Text:** "Determine the coordinates of the point $P$ on the line $y=-x+6$ such that $P$ is equidistant from the points $A(10,-10)$ and $O(0,0)$ (that is, so that $PA=PO$). Express your answer as an ordered pair $(a,b)$." ### Region 2: Main Body (Step-by-Step Solution) This section consists of a dialogue-style derivation. | Step | Transcribed Text | Score Marker | | :--- | :--- | :--- | | 1 | So we're given the line $y=-x+6$ and we need to find the point on that line where the line segment from that point to the point with coordinates (10,-10) is the same length as the line segment from the point to the origin. | [score = 0.01] | | 2 | Right. So one way to do that is to use the distance formula. Let's find the distance between P and (10,-10) and the distance between P and the origin. | [score = 0.01] | | 3 | Right. So the distance between P and (10,-10) will be the square root of $(x_P-10)^2+(y_P+10)^2$. Similarly the distance between P and the origin will be the square root of $(x_P)^2+(y_P)^2$. | [score = 0.01] | | 4 | So the two distances must be equal. So $(x_P-10)^2+(y_P+10)^2=(x_P)^2+(y_P)^2$. Expanding the left side gives $x_P^2-20x_P+100+y_P^2+20y_P+100=x_P^2+y_P^2$. | [score = 0.02] | | 5 | So now we can subtract the $x_P^2$ from both sides to get $-20x_P+20y_P+200=0$. | [score = 0.02] | | 6 | But we know that $y_P=-x_P+6$. So we can substitute that value in to get $-20x_P+20(-x_P+6)+200=0$. | [score = 0.03] | | 7 | Then we can distribute the 20 to get $-20x_P-20x_P+120+200=0$. Now we can combine the terms to get $-40x_P+320=0$. Then we can subtract 320 from both sides to get $-40x_P=-320$. Divide both sides by -40 to get $x_P=8$. | [score = 0.12] | | 8 | We know that $y_P=-x_P+6$ and $x_P=8$, so if we substitute $8$ for $x_P$ we get $y_P=-8+6$. | [score = 0.22] | | 9 | Right. Plugging in the values for $x_P$ and $y_P$ we found into the equation $y=-x+6$, we get $y=-8+6$, which simplifies to $y=-2$. | [score = 0.24] | | 10 | So the point $(8,-2)$ is equidistant from the given points. | [score = 0.36] | | 11 | That's correct. | [score = 0.35] | ### Region 3: Footer (Final Answer) **Text:** "#### (8,-2) [score = 0.39]" --- ## 3. Mathematical Logic Summary The solution follows a standard algebraic approach to find a point $P(x, y)$ satisfying two conditions: 1. **Constraint:** $P$ lies on the line $y = -x + 6$. 2. **Equidistance:** The distance $PA$ equals $PO$. **Key Equations Derived:** * **Distance Equality:** $(x-10)^2 + (y+10)^2 = x^2 + y^2$ * **Simplified Linear Equation:** $-20x + 20y + 200 = 0$ (or $y - x + 10 = 0$) * **Substitution:** Substituting $y = -x + 6$ into the simplified equation leads to $-40x + 320 = 0$. * **Result:** $x = 8$, which yields $y = -2$. **Final Coordinates:** $(8, -2)$

# Coordinate Geometry Problem Solution ## Problem Statement Determine the coordinates of the point **P** on the line **y = -x + 6** such that **P** is equidistant from the points **A(10, -10)** and **O(0, 0)**. Express your answer as an ordered pair **(a, b)**. --- ## Solution Process ### Step 1: Define Point P Since **P** lies on the line **y = -x + 6**, its coordinates can be expressed as: $$ P = (x, -x + 6) $$ ### Step 2: Distance Formula The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ #### Distance from P to A: $$ PA = \sqrt{(x - 10)^2 + (-x + 6 + 10)^2} = \sqrt{(x - 10)^2 + (-x + 16)^2} $$ #### Distance from P to O: $$ PO = \sqrt{(x - 0)^2 + (-x + 6 - 0)^2} = \sqrt{x^2 + (-x + 6)^2} $$ ### Step 3: Equate Distances Set $PA = PO$ and square both sides to eliminate radicals: $$ (x - 10)^2 + (-x + 16)^2 = x^2 + (-x + 6)^2 $$ ### Step 4: Expand and Simplify Expand all terms: $$ (x^2 - 20x + 100) + (x^2 - 32x + 256) = x^2 + (x^2 - 12x + 36) $$ Combine like terms: $$ 2x^2 - 52x + 356 = 2x^2 - 12x + 36 $$ Subtract $2x^2 - 12x + 36$ from both sides: $$ -40x + 320 = 0 $$ Solve for $x$: $$ x = 8 $$ ### Step 5: Find Corresponding y-Coordinate Substitute $x = 8$ into $y = -x + 6$: $$ y = -8 + 6 = -2 $$ ### Final Answer The coordinates of point **P** are: $$ \boxed{(8, -2)} $$ --- ## Verification - **PA**: $\sqrt{(8 - 10)^2 + (-2 + 10)^2} = \sqrt{4 + 64} = \sqrt{68}$ - **PO**: $\sqrt{8^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68}$ Both distances are equal, confirming the solution is correct.