Straight Lines
Equation of a Line
Grade None
Question:
<p>Let \(PS\) be the median of the triangle with vertices \(P(2, 2)\), \(Q(6, -1)\) and \(R(7, 3)\). The equation of the line passing through \((1, -1)\) and parallel to \(PS\) is</p>
<p>\(4x + 7y + 3 = 0\)</p>
<p>\(2x - 9y - 11 = 0\)</p>
<p>\(4x - 7y - 11 = 0\)</p>
<p>\(2x + 9y + 7 = 0\)</p>
Step-by-Step Solution
Key Concept: Find the midpoint S of QR, then calculate the slope of PS. Use this slope with point (1,-1) to write the parallel line equation.
<p><strong>Step 1:</strong> Find midpoint S of side QR.</p><p>S = ((6+7)/2, (-1+3)/2) = (13/2, 1)</p><p><strong>Step 2:</strong> Find slope of PS.</p><p>Slope of PS = (1 - 2)/(13/2 - 2) = -1/(9/2) = -2/9</p><p><strong>Step 3:</strong> The required line passes through (1, -1) and has slope -2/9 (parallel to PS).</p><p>Using point-slope form: y - (-1) = -2/9(x - 1)</p><p>y + 1 = -2/9(x - 1)</p><p>9(y + 1) = -2(x - 1)</p><p>9y + 9 = -2x + 2</p><p>2x + 9y + 7 = 0</p><p>∴ <strong>Answer: D</strong> (equation: 2x + 9y + 7 = 0)</p>
Correct Answer: D