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Zorluk: OrtaCoordinate Geometry and Lines

In the xyxy-plane, line LL passes through the origin (0,0)(0, 0) and the point P(a,b)P(a, b), where a>0a > 0 and b>0b > 0. Line MM is perpendicular to line LL at point PP. If line MM has a yy-intercept at (0,10)(0, 10) and b=2b = 2, what is the value of aa?

  1. A
    222\sqrt{2}
  2. 44Cevap
  3. C
    424\sqrt{2}
  4. D
    88
  5. E
    1616

Cevap

The value of aa is 44.
Line LL connects (0,0)(0,0) to (a,2)(a,2), giving it a slope of 2a\frac{2}{a}. Because line MM is perpendicular to line LL, its slope must be the negative reciprocal, −a2-\frac{a}{2}. Line MM also connects (a,2)(a,2) to its yy-intercept (0,10)(0,10), so its slope can independently be written as 10−20−a=−8a\frac{10-2}{0-a} = -\frac{8}{a}. Setting −a2=−8a-\frac{a}{2} = -\frac{8}{a} yields a2=16a^2 = 16. Since a>0a > 0, a=4a = 4.

Adım Adım Çözüm

1
Find the slope of line LL
Since line LL passes through (0,0)(0, 0) and P(a,2)P(a, 2), its slope is mL=2−0a−0=2am_L = \frac{2 - 0}{a - 0} = \frac{2}{a}.
The slope of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}.
2
Determine the slope of perpendicular line MM
The slope of line MM is mM=−1mL=−a2m_M = -\frac{1}{m_L} = -\frac{a}{2}.
Perpendicular lines have negative reciprocal slopes.
3
Calculate the slope of line MM using given points (a,2)(a, 2) and (0,10)(0, 10)
Using the slope formula: mM=10−20−a=8−a=−8am_M = \frac{10 - 2}{0 - a} = \frac{8}{-a} = -\frac{8}{a}.
Line MM passes through the point of intersection P(a,2)P(a, 2) and its yy-intercept (0,10)(0, 10).
4
Equate the two slope expressions and solve for aa
-\frac{a}{2} = -\frac{8}{a} \implies a^2 = 16 \implies a = 4 (since (since a > 0$).
Both expressions represent the slope of line MM.

Anahtar Kavram

Perpendicular Slopes and Line Equations in Coordinate Geometry
Tahmini Süre:1m 30s
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Konular

Coordinate Geometry and Lines

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