Question

If \( \triangle ABC \) is an isosceles triangle and the coordinates of the base points are \( B(1, 3) \) and \( C(-2, 7) \), the coordinates of \( A \) can be:

• A. \( (1, 6) \)
• B. \( \left(-\frac{1}{8}, 5\right) \)
• C. \( \left(\frac{5}{6}, 6\right) \)
• D. \( \left(-7, -\frac{1}{8}\right) \)

Hint:

For isosceles triangles, use the midpoint and perpendicular bisector properties to locate the third vertex

Answer 1

The Correct Option is C, D

1. The midpoint *M* of *BC* is calculated as:

\[M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{1 - 2}{2}, \frac{3 + 7}{2}\right) = \left(-\frac{1}{2}, 5\right).\]

This midpoint is the symmetry point for the isosceles triangle.

2. The slope of *BC* is:

\[m_{BC} = \frac{7 - 3}{-2 - 1} = -\frac{4}{3}.\]

3. The slope of the perpendicular bisector is the negative reciprocal of *BC*'s slope:

\[m_\text{perp} = \frac{3}{4}.\]

4. The perpendicular bisector's equation, passing through \(M\left(-\frac{1}{2}, 5\right)\), is:

\[y - 5 = \frac{3}{4}\left(x + \frac{1}{2}\right).\]

Simplified:

\[y - 5 = \frac{3}{4}x + \frac{3}{8} \Rightarrow y = \frac{3}{4}x + \frac{43}{8}.\]

5. Vertex *A* lies on this perpendicular bisector, and distances from *B* and *C* must be equal.

6. Using the distance formula between \(A(x,y)\) and \(B(1,3)\)

\[d_{AB} = \sqrt{(x-1)^2 + (y-3)^2}.\]

The distance \(d_{AC}\) between \(A(x,y)\) and \(C(-2,7)\) is:

\[d_{AC} = \sqrt{(x+2)^2 + (y-7)^2}.\]

7. Equate \(d_{AB} = d_{AC}\):

\[\sqrt{(x-1)^2 + (y-3)^2} = \sqrt{(x+2)^2 + (y-7)^2}.\]

8. Square both sides and simplify:

\[(x-1)^2 + (y-3)^2 = (x+2)^2 + (y-7)^2.\]

Expanded:

\[x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + 4x + 4 + y^2 - 14y + 49.\]

Simplified:

\[-6x + 10 - 8y + 40 = 0 \Rightarrow -6x + 8y + 10 = 0 \Rightarrow -6x + 8y - 25 = 0.\]

9. Solve this equation along with the perpendicular bisector equation to find \(A(x,y)\). The possible coordinates of *A* are: \(\left(-\frac{5}{6}, 6\right)\), \((-7, -\frac{1}{8})\)

Therefore, the correct answers are (C) and (D).