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Zorluk: OrtaCoordinate Geometry of Straight Lines

A straight line L1L_1 passes through the points (k,2)(k, 2) and (4,8)(4, 8). If L1L_1 is perpendicular to the line 2x−3y+6=02x - 3y + 6 = 0, what is the value of kk?

  1. A
    -5
  2. B
    0
  3. 8Cevap
  4. D
    13

Cevap

The value of kk is 88.
The line 2x−3y+6=02x - 3y + 6 = 0 has a gradient of 23\frac{2}{3}. A line perpendicular to it must have a gradient of −32-\frac{3}{2}. Calculating the gradient of L1L_1 using the points (k,2)(k, 2) and (4,8)(4, 8) gives 8−24−k=64−k\frac{8-2}{4-k} = \frac{6}{4-k}. Equating 64−k=−32\frac{6}{4-k} = -\frac{3}{2} and solving for kk gives k=8k = 8.

Adım Adım Çözüm

1
Find the gradient of the given line 2x−3y+6=02x - 3y + 6 = 0
Rearranging into y=mx+cy = mx + c form gives 3y=2x+6  ⟹  y=23x+23y = 2x + 6 \implies y = \frac{2}{3}x + 2. Therefore, the gradient m2=23m_2 = \frac{2}{3}.
To find the perpendicular gradient, we first need the gradient of the given line.
2
Determine the gradient of line L1L_1
Since L1L_1 is perpendicular to the given line, its gradient m1m_1 satisfies m1⋅m2=−1m_1 \cdot m_2 = -1. Thus, m1=−12/3=−32m_1 = -\frac{1}{2/3} = -\frac{3}{2}.
Perpendicular lines have gradients that are negative reciprocals of each other.
3
Express the gradient of L1L_1 using the coordinates (k,2)(k, 2) and (4,8)(4, 8) and solve for kk
Gradient formula: m1=y2−y1x2−x1=8−24−k=64−km_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 2}{4 - k} = \frac{6}{4 - k}. Setting this equal to −32-\frac{3}{2}:
64−k=−32\frac{6}{4 - k} = -\frac{3}{2}
6⋅2=−3(4−k)6 \cdot 2 = -3(4 - k)
12=−12+3k12 = -12 + 3k
3k=24  ⟹  k=83k = 24 \implies k = 8.
Equating the slope calculated from points to the perpendicular slope allows solving for the unknown coordinate kk.

Anahtar Kavram

Perpendicular gradients and line slope formula
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Konular

Coordinate Geometry of Straight Lines

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