Straight Lines
Coordinate Geometry Basics
Grade 11
Question:
<p>ABC is an isosceles triangle. If the coordinates of the base are B(1, 3) and C(-2, 7), the coordinates of vertex A is</p>
<p>(a) $\left(-\frac{1}{2}, -5\right)$</p>
<p>(b) $(1, 6)$</p>
<p>(c) $\left(\frac{5}{6}, 6\right)$</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the distance formula and the isosceles condition (equal sides) to set up an equation relating the coordinates of vertex A.
<p><strong>Solution:</strong> Let coordinates of vertex A be $(x, y)$.</p><p>Using the distance formula:</p><p>$$AB = \sqrt{(x-1)^2 + (y-3)^2}$$</p><p>$$AC = \sqrt{(x+2)^2 + (y-7)^2}$$</p><p>Since triangle ABC is isosceles, $AB^2 = AC^2$:</p><p>$$(x-1)^2 + (y-3)^2 = (x+2)^2 + (y-7)^2$$</p><p>Expanding:</p><p>$$x^2 + 1 - 2x + y^2 + 9 - 6y = x^2 + 4 + 4x + y^2 + 49 - 14y$$</p><p>Simplifying:</p><p>$$6x - 8y + 43 = 0$$</p><p>Checking option (c): $6 \cdot \frac{5}{6} - 8 \cdot 6 + 43 = 5 - 48 + 43 = 0$ ✓</p><p><strong>Answer: (c)</strong></p>
Correct Answer: C