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Difficulty: MediumCoordinate Geometry of Straight Lines

A straight line LL passes through the point (4,−2)(4, -2) and is perpendicular to the line defined by the equation 3x−2y+8=03x - 2y + 8 = 0. What is the yy-intercept of line LL?

  1. 23\frac{2}{3}Answer
  2. B
    −8-8
  3. C
    44
  4. D
    −143-\frac{14}{3}

Answer

The yy-intercept of line LL is 23\frac{2}{3}.
The given line 3x−2y+8=03x - 2y + 8 = 0 has a gradient of 32\frac{3}{2}. A perpendicular line must have a gradient equal to the negative reciprocal, which is −23-\frac{2}{3}. Substituting the point (4,−2)(4, -2) into y−y1=m(x−x1)y - y_1 = m(x - x_1) gives y+2=−23(x−4)y + 2 = -\frac{2}{3}(x - 4), which simplifies to y=−23x+23y = -\frac{2}{3}x + \frac{2}{3}. Setting x=0x = 0 yields the yy-intercept of 23\frac{2}{3}.

Step-by-Step Solution

1
Find the gradient of the given line.
Expressing 3x−2y+8=03x - 2y + 8 = 0 in slope-intercept form y=mx+cy = mx + c gives 2y=3x+8  ⟹  y=32x+42y = 3x + 8 \implies y = \frac{3}{2}x + 4. The gradient m1=32m_1 = \frac{3}{2}.
The gradient of the given line is required to determine the slope of line LL.
2
Determine the gradient of line LL.
Since line LL is perpendicular to the given line, mL=−1m1=−13/2=−23m_L = -\frac{1}{m_1} = -\frac{1}{3/2} = -\frac{2}{3}.
Perpendicular lines have gradients whose product is −1-1 (m1⋅mL=−1m_1 \cdot m_L = -1).
3
Find the equation of line LL using point-slope form.
Using (x1,y1)=(4,−2)(x_1, y_1) = (4, -2) and mL=−23m_L = -\frac{2}{3}:
y−y1=mL(x−x1)y - y_1 = m_L(x - x_1)
y−(−2)=−23(x−4)y - (-2) = -\frac{2}{3}(x - 4)
y+2=−23x+83y + 2 = -\frac{2}{3}x + \frac{8}{3}
y=−23x+83−2y = -\frac{2}{3}x + \frac{8}{3} - 2
y=−23x+23y = -\frac{2}{3}x + \frac{2}{3}
To find the yy-intercept, we need the complete equation of line LL.
4
Identify the yy-intercept.
Comparing y=−23x+23y = -\frac{2}{3}x + \frac{2}{3} to y=mx+cy = mx + c, the yy-intercept c=23c = \frac{2}{3}.
The constant term cc in y=mx+cy = mx + c represents the yy-intercept.

Key Concept

Perpendicular lines in coordinate geometry have gradients that are negative reciprocals of each other (m1⋅m2=−1m_1 \cdot m_2 = -1).
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Topics

Coordinate Geometry of Straight Lines

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