Question

The equations of the perpendicular bisectors of the sides AB and AC of \( \triangle ABC \) are \( x - y + 5 = 0 \) and \( x + 2y = 0 \) respectively. If the coordinates of \( A \) are \( (1, -2) \), then the equation of the line BC is:

• A. \( 14x + 23y - 40 = 0 \)
• B. \( 13x - 9y - 14 = 0 \)
• C. \( 9x - 14y - 25 = 0 \)
• D. \( 8x + 15y - 30 = 0 \)

Hint:

The perpendicular bisector of a segment is the line that passes through the midpoint of the segment, and is perpendicular to the segment.

Answer 1

The Correct Option is A

Let \(B = (h,k)\).The midpoint of \(AB\) is \(\left( \frac{h + 1}{2}, \frac{k
- 2}{2} \right)\), and this point lies on the line \(x
- y + 5 = 0\), so \[\frac{h + 1}{2}
- \frac{k
- 2}{2} + 5 = 0.\]This simplifies to \(h
- k + 13 = 0\). [asy] unitsize(2 cm); pair A, B, C, D, E; A = (1,
-2); B = (2,15); C = (8,
-4); D = (A + B)/2; E = (A + C)/2; draw(A
-
-B
-
-C
-
-cycle); draw(D
-
-D + 3*(D
- E)); draw(E
-
-E + 3*(E
- D)); label("$A$", A, SE); label("$B$", B, N); label("$C$", C, SE); label("$D$", D, NE); label("$E$", E, S); [/asy] Also, the slope of \(AB\) is \(\frac{k + 2}{h
- 1}\), and the slope of the line \(x
- y + 5 = 0\) is 1, so \[\frac{k + 2}{h
- 1} \cdot 1 =
-1.\]This simplifies to \(h + k + 1 = 0\). Solving the system \(h
- k + 13 = 0\) and \(h + k + 1 = 0\), we find \((h,k) = (
-7,6)\). Let \(C = (p,q)\).The midpoint of \(AC\) is \(\left( \frac{p + 1}{2}, \frac{q
- 2}{2} \right)\), and this point lies on the line \(x + 2y = 0\), so \[\frac{p + 1}{2} + 2 \cdot \frac{q
- 2}{2} = 0.\]This simplifies to \(p + 2q
- 3 = 0\). Also, the slope of \(AC\) is \(\frac{q + 2}{p
- 1}\), and the slope of the line \(x + 2y = 0\) is \(
-\frac{1}{2}\), so \[\frac{q + 2}{p
- 1} \cdot
-\frac{1}{2} =
-1.\]This simplifies to \(p
- 2q
- 5 = 0\). Solving the system \(p + 2q
- 3 = 0\) and \(p
- 2q
- 5 = 0\), we find \((p,q) = (4,
-1)\). Then the slope of \(BC\) is \[\frac{6
- (
-1)}{
-7
- 4} =
-\frac{7}{11},\]so the equation of line \(BC\) is of the form \[y
- 6 =
-\frac{7}{11} (x + 7),\]which simplifies to \(14x + 23y
- 40 = 0\).