Question

Difficulty: MediumCoordinate Geometry of Straight Lines

If the line 3x+py7=03x + py - 7 = 0 is perpendicular to the line passing through the points (1,2)(1, -2) and (4,7)(4, 7), what is the value of pp?

  1. 99Answer
  2. B
    1-1
  3. C
    9-9
  4. D
    33

Answer

The value of pp is 99.
The line passing through (1,2)(1, -2) and (4,7)(4, 7) has a gradient of m1=7(2)41=3m_1 = \frac{7 - (-2)}{4 - 1} = 3. The equation 3x+py7=03x + py - 7 = 0 can be rewritten as y=3px+7py = -\frac{3}{p}x + \frac{7}{p}, giving a gradient of m2=3pm_2 = -\frac{3}{p}. For perpendicular lines, the product of their gradients must equal 1-1, so 3×(3p)=13 \times \left(-\frac{3}{p}\right) = -1, which simplifies to p=9p = 9.

Step-by-Step Solution

1
Calculate the gradient (m1m_1) of the line passing through (1,2)(1, -2) and (4,7)(4, 7)
m1=7(2)41=93=3m_1 = \frac{7 - (-2)}{4 - 1} = \frac{9}{3} = 3
The gradient between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by y2y1x2x1\frac{y_2 - y_1}{x_2 - x_1}.
2
Express the line 3x+py7=03x + py - 7 = 0 in slope-intercept form (y=mx+cy = mx + c) to find its gradient (m2m_2)
py=3x+7    y=3px+7ppy = -3x + 7 \implies y = -\frac{3}{p}x + \frac{7}{p}, so m2=3pm_2 = -\frac{3}{p}
The coefficient of xx when solved for yy represents the gradient of the straight line.
3
Apply the perpendicularity condition m1m2=1m_1 \cdot m_2 = -1 and solve for pp
3(3p)=1    9p=1    p=93 \cdot \left(-\frac{3}{p}\right) = -1 \implies -\frac{9}{p} = -1 \implies p = 9
Two non-vertical lines are perpendicular if and only if the product of their gradients is 1-1.

Key Concept

Perpendicular Lines and Gradients
Estimated Time:1m 30s
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