Question

Difficulty: HardCoordinate Geometry of Straight Lines

A straight line LL has a yy-intercept of 4-4 and is perpendicular to the line segment connecting the points P(3,1)P(-3, 1) and Q(5,5)Q(5, 5). If the line LL intersects the xx-axis at (a,0)(a, 0), what is the value of aa?

  1. A
    8-8
  2. 2-2Answer
  3. C
    22
  4. D
    88

Answer

The value of aa is 2-2.
The gradient of PQPQ is calculated as m1=515(3)=12m_1 = \frac{5 - 1}{5 - (-3)} = \frac{1}{2}. Because line LL is perpendicular to PQPQ, its slope is m2=2m_2 = -2. Combining this with the given yy-intercept of 4-4 yields the equation y=2x4y = -2x - 4. Setting y=0y = 0 to determine the xx-intercept gives 0=2a40 = -2a - 4, which solves to a=2a = -2.

Step-by-Step Solution

1
Find the gradient m1m_1 of the line segment PQPQ
m1=515(3)=48=12m_1 = \frac{5 - 1}{5 - (-3)} = \frac{4}{8} = \frac{1}{2}
The formula for the gradient between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
2
Calculate the gradient m2m_2 of line LL
m2=1m1=11/2=2m_2 = -\frac{1}{m_1} = -\frac{1}{1/2} = -2
Perpendicular lines have gradients satisfying m1m2=1m_1 \cdot m_2 = -1.
3
Determine the equation of line LL
y=2x4y = -2x - 4
Using the slope-intercept form y=mx+cy = mx + c, where m=2m = -2 and c=4c = -4.
4
Find the xx-intercept by substituting y=0y = 0 and x=ax = a
0=2a4    2a=4    a=20 = -2a - 4 \implies 2a = -4 \implies a = -2
At the xx-axis, the yy-coordinate is always equal to 00.

Key Concept

Perpendicular Line Slopes and Coordinate Intercepts
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