Straight Lines
Distance from a line
Grade 11
Question:
<p><strong>948.</strong> Let \(A(x_1, y_1)\), \(B(x_2, y_2)\) and \(C(x_3, y_3)\) be the vertices of a triangle such that algebraic sum of perpendicular distance from \(A\), \(B\) and \(C\) to the variable line \(ax + by + c = 0\) is always '0'. If \(3a + 2b + c = 0\), then find the value of \(\displaystyle\sum_{i=1}^{3}(x_i + y_i)\).</p>
Step-by-Step Solution
Key Concept: If the algebraic sum of perpendicular distances from three points to a variable line is always zero, then the centroid of the triangle must lie on every such line. This means the line must pass through the centroid, so the centroid coordinates satisfy the constraint equation.
<p><strong>Step 1:</strong> The perpendicular distance from point $(x_i, y_i)$ to line $ax + by + c = 0$ is $\frac{ax_i + by_i + c}{\sqrt{a^2 + b^2}}$ (algebraic/signed distance).</p><p><strong>Step 2:</strong> Given condition: $\frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} + \frac{ax_2 + by_2 + c}{\sqrt{a^2 + b^2}} + \frac{ax_3 + by_3 + c}{\sqrt{a^2 + b^2}} = 0$</p><p><strong>Step 3:</strong> Simplifying: $a(x_1 + x_2 + x_3) + b(y_1 + y_2 + y_3) + 3c = 0$ for all values of $a, b, c$ satisfying the constraint.</p><p><strong>Step 4:</strong> The centroid $G$ is $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$. Let $S_x = x_1 + x_2 + x_3$ and $S_y = y_1 + y_2 + y_3$.</p><p><strong>Step 5:</strong> From Step 3: $aS_x + bS_y + 3c = 0$ for all $(a,b,c)$ with $3a + 2b + c = 0$.</p><p><strong>Step 6:</strong> Using the constraint $c = -3a - 2b$, substitute: $aS_x + bS_y + 3(-3a - 2b) = 0$</p><p>$aS_x + bS_y - 9a - 6b = 0$</p><p>$a(S_x - 9) + b(S_y - 6) = 0$ for all $a, b$.</p><p><strong>Step 7:</strong> For this to hold for all values of $a$ and $b$: $S_x - 9 = 0$ and $S_y - 6 = 0$</p><p>Therefore: $S_x = 9$ and $S_y = 6$</p><p><strong>Step 8:</strong> $\sum_{i=1}^{3}(x_i + y_i) = (x_1 + x_2 + x_3) + (y_1 + y_2 + y_3) = 9 + 6 = 15$</p><p>∴ Answer: <strong>15</strong></p>
Correct Answer: 15