Straight Lines
Isosceles triangle and line equations
Grade 11
Question:
<p>Let \(\triangle ABC\) be an isosceles triangle with \(AB = AC\). If \(AB: 4x + y = 7\), \(AC: x + 4y = 7\) and \(BC\) is passing through \((1, 1)\), then possible equation of \(BC\) is:</p>
<p>\(3x + 2y = 5\)</p>
<p>\(x + y = 2\)</p>
<p>\(2x + 3y = 5\)</p>
<p>\(x - y = 0\)</p>
Step-by-Step Solution
Key Concept: In an isosceles triangle with AB = AC, the angle bisector from A is perpendicular to BC. Use the angle bisector property: the angle bisector from A bisects the angle between lines AB and AC, then BC is perpendicular to this bisector.
<p><strong>Step 1:</strong> Find the angle bisectors of lines AB: 4x + y = 7 and AC: x + 4y = 7.</p><p>The angle bisectors are given by: $\frac{4x + y - 7}{\sqrt{16+1}} = \pm \frac{x + 4y - 7}{\sqrt{1+16}}$</p><p>This gives: $\frac{4x + y - 7}{\sqrt{17}} = \pm \frac{x + 4y - 7}{\sqrt{17}}$</p><p><strong>Step 2:</strong> Taking the positive sign: 4x + y - 7 = x + 4y - 7 → 3x - 3y = 0 → x = y (angle bisector from A)</p><p>Taking the negative sign: 4x + y - 7 = -(x + 4y - 7) → 5x + 5y = 14 → x + y = 14/5 (other bisector)</p><p><strong>Step 3:</strong> Since AB = AC, the line BC is perpendicular to the angle bisector from vertex A. The angle bisector from A has slope 1 (from x = y).</p><p><strong>Step 4:</strong> A line perpendicular to slope 1 has slope -1. So BC has the form: y - 1 = -1(x - 1) → y - 1 = -x + 1 → x + y = 2</p><p><strong>Step 5:</strong> Verify: BC passes through (1,1): 1 + 1 = 2 ✓</p><p>∴ Answer: B (equation is x + y = 2 or equivalent form)</p>
Correct Answer: B