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Zorluk: KolayCoordinate Geometry of Straight Lines

What is the gradient of a straight line that is perpendicular to the line passing through the points (1,−2)(1, -2) and (5,6)(5, 6)?

  1. −12-\frac{1}{2}Cevap
  2. B
    22
  3. C
    −2-2
  4. D
    12\frac{1}{2}

Cevap

−12-\frac{1}{2}
The slope of the given line segment is computed as 6−(−2)5−1=2\frac{6 - (-2)}{5 - 1} = 2. The perpendicular gradient is the negative reciprocal of 22, which gives −12-\frac{1}{2}.

Adım Adım Çözüm

1
Calculate the gradient of the line passing through the two points (1,−2)(1, -2) and (5,6)(5, 6).
Using m1=y2−y1x2−x1m_1 = \frac{y_2 - y_1}{x_2 - x_1}, we obtain m1=6−(−2)5−1=84=2m_1 = \frac{6 - (-2)}{5 - 1} = \frac{8}{4} = 2.
The slope of a straight line through two points is defined as the change in yy divided by the change in xx.
2
Apply the perpendicularity condition to find the perpendicular slope m2m_2.
Since m1⋅m2=−1m_1 \cdot m_2 = -1, m2=−1m1=−12m_2 = -\frac{1}{m_1} = -\frac{1}{2}.
Two non-vertical lines are perpendicular if and only if the product of their gradients is −1-1.

Anahtar Kavram

Perpendicular Line Gradients
Tahmini Süre:45s
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