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Zorluk: OrtaCoordinate Geometry and Lines

In the xyxy-plane, line L1L_1 is defined by the equation y=3x−4y = 3x - 4. Line L2L_2 is perpendicular to line L1L_1 and passes through the point (6,2)(6, 2). What is the xx-intercept of line L2L_2?

Cevap: 12

Cevap

12
Line L1L_1 has a slope of 33. A line perpendicular to L1L_1 must have a slope equal to the negative reciprocal of 33, which is −13-\frac{1}{3}. Substituting the point (6,2)(6, 2) into the point-slope form gives y−2=−13(x−6)y - 2 = -\frac{1}{3}(x - 6), which simplifies to y=−13x+4y = -\frac{1}{3}x + 4. Setting y=0y = 0 yields 0=−13x+40 = -\frac{1}{3}x + 4, giving x=12x = 12 as the xx-intercept.

Adım Adım Çözüm

1
Find the slope of line L1L_1
Slope m1=3m_1 = 3
The equation y=3x−4y = 3x - 4 is in slope-intercept form y=mx+by = mx + b, where m=3m = 3.
2
Determine the slope of perpendicular line L2L_2
Slope m2=−13m_2 = -\frac{1}{3}
Perpendicular lines have slopes that are negative reciprocals of each other (m1⋅m2=−1m_1 \cdot m_2 = -1).
3
Find the equation of line L2L_2
y=−13x+4y = -\frac{1}{3}x + 4
Substitute point (6,2)(6, 2) into point-slope formula y−2=−13(x−6)y - 2 = -\frac{1}{3}(x - 6).
4
Calculate the xx-intercept of line L2L_2
x=12x = 12
Set y=0y = 0 in the line equation and solve for xx: 0=−13x+4  ⟹  x=120 = -\frac{1}{3}x + 4 \implies x = 12.

Anahtar Kavram

Perpendicular lines have negative reciprocal slopes (m1⋅m2=−1m_1 \cdot m_2 = -1). The xx-intercept is the point where y=0y = 0.
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Konular

Coordinate Geometry and Lines

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