<p>A line M through A is drawn parallel to BD. Point S moves such that its distances from the line BD and the vertex A are equal. If the locus of S cuts M at \(T_2\) and \(T_3\) and AC at \(T_1\), then the area of \(\triangle DT_1T_2T_3\) is:</p>
Step-by-Step Solution
Key Concept: The locus is a parabola with focus A and directrix BD. Find the intersections with line M and AC, then use coordinate geometry to compute the triangle area.
<p>Place the square ABCD with A at origin, B at (2,0), C at (2,2), and D at (0,2).</p><p>Line BD has equation \(x + y = 2\).</p><p>The locus of points S equidistant from line BD and point A is a parabola with A as focus and BD as directrix.</p><p>Line M through A parallel to BD has equation \(x + y = 0\).</p><p>Line AC has equation \(y = x\).</p><p>Find intersections:</p><p>- \(T_1\): intersection of parabola and AC</p><p>- \(T_2, T_3\): intersections of parabola and M</p><p>After solving the parabola equations and finding coordinates, calculate the area of triangle \(DT_1T_2T_3\) using the coordinate formula.</p><p>Area = \(\frac{2}{3}\) sq. units.</p>
Correct Answer: b