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Zorluk: OrtaCoordinate Geometry of Straight Lines

A straight line LL passes through the point (2,−3)(2, -3) and is perpendicular to the line 4x+5y−20=04x + 5y - 20 = 0. Calculate the xx-intercept of line LL.

Cevap: 4.4

Cevap

The xx-intercept of line LL is 4.44.4 (or 225\frac{22}{5}).
The given line 4x+5y−20=04x + 5y - 20 = 0 has a slope of −45-\frac{4}{5}. The line perpendicular to it must have a slope equal to the negative reciprocal, which is 54\frac{5}{4}. Using the point-slope formula with point (2,−3)(2, -3), the equation of line LL is y+3=54(x−2)y + 3 = \frac{5}{4}(x - 2), simplifying to y=54x−112y = \frac{5}{4}x - \frac{11}{2}. Setting y=0y = 0 yields 54x=112\frac{5}{4}x = \frac{11}{2}, giving an xx-intercept of x=225=4.4x = \frac{22}{5} = 4.4.

Adım Adım Çözüm

1
Determine the gradient of the given line 4x+5y−20=04x + 5y - 20 = 0
Gradient m1=−45m_1 = -\frac{4}{5}
Converting to slope-intercept form y=−45x+4y = -\frac{4}{5}x + 4 reveals the slope.
2
Calculate the perpendicular gradient for line LL
Gradient m=54m = \frac{5}{4}
Perpendicular lines have negative reciprocal gradients (m1⋅m2=−1m_1 \cdot m_2 = -1).
3
Derive the equation of line LL using point (2,−3)(2, -3)
y=54x−112y = \frac{5}{4}x - \frac{11}{2}
Substitute the point (2,−3)(2, -3) and gradient m=54m = \frac{5}{4} into point-slope form.
4
Solve for the xx-intercept by setting y=0y = 0
x=4.4x = 4.4
The xx-intercept is defined as the point where the line crosses the xx-axis (y=0y = 0).

Anahtar Kavram

Perpendicular line slope relationships and x-intercept calculations
Tahmini Süre:1m 30s
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Konular

Coordinate Geometry of Straight Lines

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