Question

determine the coordinates of point k so that quadrilateral abdc is similar to quadrilateral efkl.

Explanation:

Step1: Identify Coordinates of Original Quadrilateral

First, find coordinates of points in \(ABDC\):
- \(A(1,2)\), \(B(1,4)\), \(C(3,2)\), \(D(2,6)\)
- \(AB\) is vertical: length \(4 - 2 = 2\)
- \(AC\) is horizontal: length \(3 - 1 = 2\)

For \(EFKL\):
- \(E(6,2)\), \(F(12,2)\), \(L(6,8)\)
- \(EF\) length: \(12 - 6 = 6\)
- \(EL\) length: \(8 - 2 = 6\)

Step2: Determine Scale Factor

Scale factor from \(ABDC\) to \(EFKL\):
\(EF/AC = 6/2 = 3\), \(EL/AB = 6/2 = 3\). So scale factor \(k = 3\).

Step3: Find Coordinates of \(K\)

In \(ABDC\), \(D\) is related to \(A,B,C\). In \(EFKL\), \(K\) should correspond to \(D\).
Vector from \(A\) to \(D\): \(D - A = (2 - 1, 6 - 2) = (1,4)\).
Scale this vector by \(k = 3\): \((1\times3, 4\times3) = (3,12)\).
Add to \(E(6,2)\): \(K = (6 + 3, 2 + 12) = (9,14)\)? Wait, no—wait, \(ABDC\) structure: \(AB\) vertical, \(AC\) horizontal, \(D\) at (2,6). For \(EFKL\), \(E(6,2)\), \(F(12,2)\) (horizontal), \(L(6,8)\) (vertical). So \(K\) should be such that \(EFKL\) is similar. Wait, maybe better: \(ABDC\) has \(A(1,2)\), \(B(1,4)\) (vertical), \(C(3,2)\) (horizontal), \(D(2,6)\) (top). So \(AB\) length 2, \(AC\) length 2, \(BD\) vector (1,2), \(CD\) vector (-1,4)? No, better to see \(ABDC\) is a quadrilateral with \(AB\) vertical, \(AC\) horizontal, then \(D\) connected to \(B\) and \(C\). So in \(EFKL\), \(E(6,2)\), \(F(12,2)\) (horizontal, length 6), \(L(6,8)\) (vertical, length 6). So \(K\) should be at \(E + (F - E)/2 + (L - E)/2\)? Wait, no—wait \(D\) is at (2,6), which is \(A + (C - A)/2 + (B - A)/2\)? \(A(1,2)\), \(C(3,2)\) midpoint (2,2), \(B(1,4)\) midpoint? No, \(D\) is (2,6). So vector from \(A\) to \(D\) is (1,4). In \(EFKL\), \(E(6,2)\), so \(K = E + (D - A) \times 3\)? Wait \(D - A = (1,4)\), scale by 3: (3,12), so \(E + (3,12) = (9,14)\)? But let's check lengths. Wait \(AB\) is 2, \(EL\) is 6 (scale 3). \(AC\) is 2, \(EF\) is 6 (scale 3). So \(D\) is at (2,6), which is \(A + (1,4)\). So \(K\) should be \(E + (1\times3, 4\times3) = (6 + 3, 2 + 12) = (9,14)\)? Wait, no, maybe I messed up. Wait \(ABDC\): \(A(1,2)\), \(B(1,4)\) (vertical, length 2), \(C(3,2)\) (horizontal, length 2), \(D(2,6)\) (so from \(A\), right 1, up 4). \(EFKL\): \(E(6,2)\), \(F(12,2)\) (horizontal, length 6), \(L(6,8)\) (vertical, length 6). So \(K\) should be \(E\) right 1*3, up 4*3? Wait 1*3=3, 4*3=12. So \(E(6,2) + (3,12) = (9,14)\)? Wait, but let's check the shape. \(ABDC\) has \(AB\) vertical, \(AC\) horizontal, \(D\) at (2,6). So \(K\) should be at (9,14)? Wait, no, maybe the x-coordinate: \(A\) is at x=1, \(C\) at x=3 (difference 2), \(E\) at x=6, \(F\) at x=12 (difference 6, scale 3). \(D\) is at x=2 (midway between 1 and 3), so \(K\) should be midway between 6 and 12 in x: (6+12)/2=9, and y: midway between 2 and 8? No, \(D\) is at y=6, which is 2 + 4, and \(L\) is at y=8, which is 2 + 6. Wait, \(AB\) length 2, \(EL\) length 6 (scale 3). \(BD\) length: distance from \(B(1,4)\) to \(D(2,6)\) is \(\sqrt{1^2 + 2^2} = \sqrt{5}\). \(KL\) should be \(\sqrt{5} \times 3 = \sqrt{45}\). Let's check \(K(9,14)\) to \(L(6,8)\): distance \(\sqrt{(9-6)^2 + (14-8)^2} = \sqrt{9 + 36} = \sqrt{45}\), correct. \(K(9,14)\) to \(F(12,2)\): distance \(\sqrt{(12-9)^2 + (2-14)^2} = \sqrt{9 + 144} = \sqrt{153}\), and \(D(2,6)\) to \(C(3,2)\): \(\sqrt{1 + 16} = \sqrt{17}\), \(\sqrt{17} \times 9 = \sqrt{153}\), correct. So \(K\) is (9,14)? Wait, no, wait the grid: y-axis up to 15, so (9,14) is on the grid. Wait, but let's re-express:

Original \(ABDC\):
- \(A(1,2)\), \(B(1,4)\) (vertical, length 2)
- \(C(3,2)\) (horizontal from A, length 2)
- \(D(2,6)\) (midway x between A and C, y above B)

Scaled \(EFKL\):
- \(E(6,2)\), \(F(12,2)\) (horizontal, length 6 = 2*3)
- \(L(6,8)\) (vertical, length 6 = 2*3)
- \(K\) should be midway x between E and F (6 and 12, midpoint x=9) and y: from E(2) up by (6 - 2)*2? Wait \(D\) is at y=6, which is 2 + 4, and \(B\) is at y=4 (2 + 2). So from \(E(2)\), up by 4*3? No, \(AB\) length 2, \(EL\) length 6 (3x). \(BD\) length: from \(B(1,4)\) to \(D(2,6)\) is (1,2), so scaled by 3: (3,6). So from \(L(6,8)\), add (3,6)? No, \(L\) is (6,8), \(F\) is (12,2). Wait, maybe \(K\) is at (9,14)? Wait, no, let's calculate the coordinates properly.

Wait, the vector from \(A\) to \(D\) is (2 - 1, 6 - 2) = (1, 4). The vector from \(E\) to \(K\) should be the same as from \(A\) to \(D\) scaled by the scale factor. The scale factor is \(EF / AC = (12 - 6)/(3 - 1) = 6/2 = 3\). So the vector from \(E\) to \(K\) is \(3 \times (1, 4) = (3, 12)\). Therefore, \(K = E + (3, 12) = (6 + 3, 2 + 12) = (9, 14)\). Wait, but let's check the y-coordinate: \(L\) is at (6,8), so \(EL\) is length 6, \(AB\) is length 2, scale 3. \(D\) is at y=6, which is 2 (A's y) + 4, so \(K\) should be 2 (E's y) + 4*3 = 14, which matches. X-coordinate: \(A\) is at x=1, \(D\) at x=2 (1 + 1), so \(E\) at x=6, \(K\) at x=6 + 1*3 = 9. Yes, that makes sense.

Answer:

\(K = (9, 14)\)