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>(a) \(3x + 2y = 5\)</p>
<p>(b) \(x + y = 2\)</p>
<p>(c) \(2x + 3y = 5\)</p>
<p>(d) \(x - y = 0\)</p>
Step-by-Step Solution
Key Concept: Since AB = AC and the triangle is isosceles, the angle bisector from vertex A is perpendicular to BC. Find this angle bisector, then BC must be perpendicular to it and pass through (1,1).
<p><strong>Step 1:</strong> Find point A (intersection of AB and AC).</p><p>Solve 4x + y = 7 and x + 4y = 7:</p><p>From first: y = 7 - 4x</p><p>Substitute: x + 4(7 - 4x) = 7 → x + 28 - 16x = 7 → -15x = -21 → x = 7/5</p><p>Then y = 7 - 4(7/5) = 7/5</p><p>So A = (7/5, 7/5)</p><p><strong>Step 2:</strong> Find the angle bisector of ∠BAC.</p><p>Slopes: m₁ = -4 (from 4x + y = 7), m₂ = -1/4 (from x + 4y = 7)</p><p>The angle bisector from A has slope m such that it bisects the angle. Using the angle bisector formula or noting the symmetry of coefficients (4 and 4 in perpendicular roles):</p><p>Angle bisector slope = 1 (since the lines are symmetrically placed with respect to y = x)</p><p><strong>Step 3:</strong> Since AB = AC, the angle bisector from A is perpendicular to BC.</p><p>If angle bisector has slope 1, then BC has slope -1.</p><p><strong>Step 4:</strong> BC passes through (1, 1) with slope -1:</p><p>y - 1 = -1(x - 1) → y - 1 = -x + 1 → x + y = 2</p><p>∴ Answer: B</p>
Correct Answer: B