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Zorluk: ZorCoordinate Geometry: Lines, Slopes, and Distance

In the xyxy-plane, line L1L_1 is defined by the equation 2x−3y=62x - 3y = 6. Line L2L_2 is perpendicular to line L1L_1 and passes through the point P(4,−1)P(4, -1). Point Q(a,b)Q(a, b) lies on line L2L_2 such that the distance between point PP and point QQ is 13\sqrt{13}. Which of the following statements regarding point QQ or line L2L_2 could be true? Select all such statements.

  1. Point QQ lies in Quadrant I.Cevap
  2. The sum of the coordinates of point QQ, a+ba + b, is equal to 44.Cevap
  3. The distance from point QQ to the origin is 2132\sqrt{13}.Cevap
  4. D
    The yy-intercept of line L2L_2 is (0,−6)(0, -6).
  5. E
    Point QQ could have coordinates (4,2)(4, 2).

Cevap

The correct statements are that point Q can lie in Quadrant I, the sum of the coordinates of point Q can equal 4, and the distance from point Q to the origin can be 2√13.
Solving the system formed by line L2L_2 (3x+2y=103x + 2y = 10) and the distance constraint (a−4)2+(b+1)2=13(a-4)^2 + (b+1)^2 = 13 yields two possible points: (2,2)(2, 2) and (6,−4)(6, -4). The point (2,2)(2, 2) lies in Quadrant I and has a coordinate sum of 2+2=42 + 2 = 4. The point (6,−4)(6, -4) has a distance to the origin of 62+(−4)2=52=213\sqrt{6^2 + (-4)^2} = \sqrt{52} = 2\sqrt{13}.

Adım Adım Çözüm

1
Determine the slope and equation of line L2L_2.
Line L1L_1 has slope m1=23m_1 = \frac{2}{3}. Therefore, perpendicular line L2L_2 has slope m2=−32m_2 = -\frac{3}{2}. Using point P(4,−1)P(4, -1), the equation of L2L_2 is y−(−1)=−32(x−4)y - (-1) = -\frac{3}{2}(x - 4), which simplifies to y=−32x+5y = -\frac{3}{2}x + 5 or 3x+2y=103x + 2y = 10.
Perpendicular lines in the coordinate plane have negative reciprocal slopes.
2
Express the distance constraint between P(4,−1)P(4, -1) and Q(a,b)Q(a, b) algebraically.
Since Q(a,b)Q(a, b) lies on L2L_2, b=−32a+5b = -\frac{3}{2}a + 5. The distance squared is (a−4)2+(b+1)2=13(a - 4)^2 + (b + 1)^2 = 13. Substituting b+1=−32(a−4)b + 1 = -\frac{3}{2}(a - 4) yields (a−4)2+(−32(a−4))2=13(a - 4)^2 + \left(-\frac{3}{2}(a - 4)\right)^2 = 13, which simplifies to 134(a−4)2=13\frac{13}{4}(a - 4)^2 = 13, so (a−4)2=4(a - 4)^2 = 4.
Applying the distance formula and substituting the line equation reduces the problem to a quadratic equation in one variable.
3
Solve for the possible coordinates of point QQ.
Taking square roots gives a−4=2a - 4 = 2 or a−4=−2a - 4 = -2, resulting in a=6a = 6 or a=2a = 2. Correspondingly, b=−4b = -4 or b=2b = 2. Thus, QQ can be (6,−4)(6, -4) or (2,2)(2, 2).
Quadratic equations of the form (x−h)2=k(x-h)^2 = k have two real solutions.
4
Evaluate each statement against the possible coordinates Q(6,−4)Q(6, -4) and Q(2,2)Q(2, 2).
For Q(2,2)Q(2, 2): it lies in Quadrant I (valid), its coordinate sum is 2+2=42 + 2 = 4 (valid). For Q(6,−4)Q(6, -4): its distance to the origin is 62+(−4)2=52=213\sqrt{6^2 + (-4)^2} = \sqrt{52} = 2\sqrt{13} (valid). The yy-intercept of L2L_2 is (0,5)(0, 5), and (4,2)(4, 2) is not on L2L_2.
Direct verification confirms which properties hold for the two solved points.

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