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Zorluk: KolaySlope of a Line

In the standard (x,y)(x, y) coordinate plane, line l1l_1 passes through the points (−2,5)(-2, 5) and (4,1)(4, 1). Line l2l_2 is perpendicular to line l1l_1. What is the slope of line l2l_2?

  1. A
    −32-\frac{3}{2}
  2. B
    −23-\frac{2}{3}
  3. C
    23\frac{2}{3}
  4. 32\frac{3}{2}Cevap
  5. E
    12\frac{1}{2}

Cevap

The slope of the perpendicular line is 32\frac{3}{2}.
To find the slope of a line perpendicular to a given line, first calculate the slope of the original line, l1l_1, using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}. Substituting the points (−2,5)(-2, 5) and (4,1)(4, 1) into the formula gives m1=1−54−(−2)=−46=−23m_1 = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}. The slope of a perpendicular line, l2l_2, is the negative reciprocal of the slope of l1l_1. The negative reciprocal of −23-\frac{2}{3} is 32\frac{3}{2}.

Adım Adım Çözüm

1
Calculate the slope of the line l1l_1 passing through the points (−2,5)(-2, 5) and (4,1)(4, 1) using the slope formula.
m1=1−54−(−2)=−46=−23m_1 = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}
The slope mm of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.
2
Find the slope of line l2l_2, which is perpendicular to l1l_1, by taking the negative reciprocal of m1m_1.
m2=−1m1=−1−23=32m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}
The product of the slopes of two perpendicular lines is −1-1, so the slope of a perpendicular line is the negative reciprocal of the original slope.

Anahtar Kavram

Slope of perpendicular lines and finding slope from two points.
Tahmini Süre:1m 0s
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Konular

Slope of a Line

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