Math Problem Statement

Dik koordinat düzleminde A(9, -1) ve B(1, 7) noktaları veriliyor. C ∈ [AB] ve |AC| = 3|BC| olduğuna göre, C noktasının koordinatlarını bulunuz.

Solution

The problem involves points A(9,1)A(9, -1) and B(1,7)B(1, 7) in the Cartesian plane. Point CC lies on the line segment ABAB and satisfies the condition AC=3BC|AC| = 3|BC|. We are tasked with finding the coordinates of point CC.


Step 1: Parametric representation of point CC

Since CC lies on the line segment ABAB, it can be represented parametrically as: C(x,y)=(1t)A+tBC(x, y) = (1-t)A + tB Where t[0,1]t \in [0, 1] and: C(x,y)=((1t)9+t1,(1t)1+t7)C(x, y) = ((1-t) \cdot 9 + t \cdot 1, (1-t) \cdot -1 + t \cdot 7) Simplify the coordinates: C(x,y)=(98t,1+8t)C(x, y) = (9 - 8t, -1 + 8t)


Step 2: Condition AC=3BC|AC| = 3|BC|

The condition AC=3BC|AC| = 3|BC| implies that the distances satisfy: AC2=9BC2|AC|^2 = 9|BC|^2 Using the distance formula, compute AC2|AC|^2 and BC2|BC|^2: AC2=(xCxA)2+(yCyA)2|AC|^2 = (x_C - x_A)^2 + (y_C - y_A)^2 BC2=(xCxB)2+(yCyB)2|BC|^2 = (x_C - x_B)^2 + (y_C - y_B)^2

Substituting A(9,1)A(9, -1), B(1,7)B(1, 7), and C(98t,1+8t)C(9 - 8t, -1 + 8t): AC2=((98t)9)2+((1+8t)(1))2=(8t)2+(8t)2=64t2+64t2=128t2|AC|^2 = ((9 - 8t) - 9)^2 + ((-1 + 8t) - (-1))^2 = (-8t)^2 + (8t)^2 = 64t^2 + 64t^2 = 128t^2 BC2=((98t)1)2+((1+8t)7)2=(88t)2+(8+8t)2=(64128t+64t2)+(64128t+64t2)|BC|^2 = ((9 - 8t) - 1)^2 + ((-1 + 8t) - 7)^2 = (8 - 8t)^2 + (-8 + 8t)^2 = (64 - 128t + 64t^2) + (64 - 128t + 64t^2) BC2=128256t+128t2|BC|^2 = 128 - 256t + 128t^2

Using the condition AC2=9BC2|AC|^2 = 9|BC|^2: 128t2=9(128256t+128t2)128t^2 = 9(128 - 256t + 128t^2) Simplify: 128t2=11522304t+1152t2128t^2 = 1152 - 2304t + 1152t^2 0=11522304t+1024t20 = 1152 - 2304t + 1024t^2 Divide by 64: 0=1836t+16t20 = 18 - 36t + 16t^2 Rearrange into standard quadratic form: 16t236t+18=016t^2 - 36t + 18 = 0 Simplify further: 8t218t+9=08t^2 - 18t + 9 = 0


Step 3: Solve the quadratic equation

Using the quadratic formula: t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Here, a=8a = 8, b=18b = -18, and c=9c = 9: t=(18)±(18)24(8)(9)2(8)t = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(8)(9)}}{2(8)} t=18±32428816t = \frac{18 \pm \sqrt{324 - 288}}{16} t=18±3616t = \frac{18 \pm \sqrt{36}}{16} t=18±616t = \frac{18 \pm 6}{16} t=2416=32ort=1216=34t = \frac{24}{16} = \frac{3}{2} \quad \text{or} \quad t = \frac{12}{16} = \frac{3}{4}

Since t[0,1]t \in [0, 1], we take t=34t = \frac{3}{4}.


Step 4: Find coordinates of CC

Substitute t=34t = \frac{3}{4} into C(x,y)=(98t,1+8t)C(x, y) = (9 - 8t, -1 + 8t): xC=9834=96=3x_C = 9 - 8 \cdot \frac{3}{4} = 9 - 6 = 3 yC=1+834=1+6=5y_C = -1 + 8 \cdot \frac{3}{4} = -1 + 6 = 5

Thus, the coordinates of CC are: C(3,5)C(3, 5)


Do you want further clarifications or details on any step? Here are some related questions:

  1. How is the parametric representation of a point on a line derived?
  2. Can we verify the solution by rechecking distances?
  3. What if the condition was AC=2BC|AC| = 2|BC|? How would it change the result?
  4. How can this problem be extended to three dimensions?
  5. Is there a geometric interpretation of AC=3BC|AC| = 3|BC|?

Tip: When solving problems involving ratios, parametric equations are often the most efficient method!

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Parametric Equations
Ratios
Distance Formula

Formulas

C(x, y) = (1-t)A + tB
Distance formula: |AB| = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Quadratic equation: ax^2 + bx + c = 0

Theorems

Ratios on a Line Segment
Quadratic Formula

Suitable Grade Level

Grades 9-12