Straight Lines
Locus
Grade 11
Question:
<p>If the equation of the locus of a point equidistant from the points \((a_1, b_1)\) and \((a_2, b_2)\) is \((a_1 - a_2)x + (b_1 - b_2)y + c = 0\), then the value of \(c\) is</p>
<p>\(\dfrac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)\)</p>
<p>\(a_1^2 + a_2^2 + b_1^2 - b_2^2\)</p>
<p>\(\dfrac{1}{2}(a_1^2 + a_2^2 - b_1^2 - b_2^2)\)</p>
<p>\(\sqrt{a_1^2 + b_1^2 - a_2^2 - b_2^2}\)</p>
Step-by-Step Solution
Key Concept: The locus of points equidistant from two points is the perpendicular bisector of the line segment joining them. Derive this by setting the distance formula equal and simplifying to find the constant term c.
<p><strong>Step 1:</strong> Set up the equidistant condition. If point P(x, y) is equidistant from A(a₁, b₁) and B(a₂, b₂):</p><p>√[(x - a₁)² + (y - b₁)²] = √[(x - a₂)² + (y - b₂)²]</p><p><strong>Step 2:</strong> Square both sides and expand:</p><p>(x - a₁)² + (y - b₁)² = (x - a₂)² + (y - b₂)²</p><p>x² - 2a₁x + a₁² + y² - 2b₁y + b₁² = x² - 2a₂x + a₂² + y² - 2b₂y + b₂²</p><p><strong>Step 3:</strong> Cancel x² and y², then rearrange:</p><p>-2a₁x + a₁² - 2b₁y + b₁² = -2a₂x + a₂² - 2b₂y + b₂²</p><p>2(a₂ - a₁)x + 2(b₂ - b₁)y = a₂² - a₁² + b₂² - b₁²</p><p><strong>Step 4:</strong> Multiply by -1 and divide by 2:</p><p>(a₁ - a₂)x + (b₁ - b₂)y = ½[(a₁² - a₂²) + (b₁² - b₂²)]</p><p><strong>Step 5:</strong> Using difference of squares and factoring:</p><p>(a₁ - a₂)x + (b₁ - b₂)y = ½(a₁ - a₂)(a₁ + a₂) + ½(b₁ - b₂)(b₁ + b₂)</p><p><strong>Step 6:</strong> Comparing with (a₁ - a₂)x + (b₁ - b₂)y + c = 0:</p><p>c = -½[(a₁ + a₂)(a₁ - a₂) + (b₁ + b₂)(b₁ - b₂)]</p><p>or equivalently: <strong>c = ½(a₂² + b₂² - a₁² - b₁²)</strong></p><p>∴ Answer: <strong>c = ½[(a₂² - a₁²) + (b₂² - b₁²)]</strong> or <strong>-½[(a₁² + b₁²) - (a₂² + b₂²)]</strong></p>
Correct Answer: A