Printable

2017 AMC 10B Problem 8

2017 AMC 10B Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

Coordinate Geometry · difficulty 2 of 5 · about 2 min · position P1-10

Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$ ?

A)
$(-8, 9)$
B)
$(-4, 8)$
C)
$(-4, 9)$
D)
$(-2, 3)$
E)
$(-1, 0)$

Answer

Key insight

In an isosceles triangle the altitude from the apex bisects the base, so D is the midpoint of BC and C = 2D - B.

Solution

Because $AB = AC$ , the altitude from the apex $A$ to base $BC$ is also the perpendicular bisector of $BC$ . Therefore $D$ is the midpoint of $\overline{BC}$ .

Going from $B$ to $D$ moves $(-1 - 2,\ 3 - (-3)) = (-3, 6)$ . Moving the same amount again from $D$ reaches $C$ :
$C = (-1 - 3,\ 3 + 6) = (-4, 9).$

Quick check: $AB^2 = 9^2 + 12^2 = 225$ and $AC^2 = 15^2 + 0^2 = 225$ , so $AB = AC$ indeed.

The answer is $\boxed{\textbf{(C)}\ (-4, 9)}$ .

Why this works

In an isosceles triangle the altitude, median, and angle bisector from the apex coincide. Translating that geometric fact into coordinates turns the problem into a midpoint computation: $C = 2D - B$ coordinate by coordinate. The equal-distance check at the end costs ten seconds and catches sign errors.

Alternative approach

Test the choices with $AC = AB = 15$ : only $(-4, 9)$ is at distance $15$ from $A(11, 9)$ , since it lies on the horizontal line $y = 9$ exactly $15$ units left of $A$ .

The trap

Averaging A and B, or reflecting B over A instead of over D.

Common mistakes

Averaging A and B, or reflecting B over A instead of over D.
Computing $D - (B - D)$ with a sign slip, landing on $(-4, -3)$ or $(2, 9)$ , neither of which is offered, and then guessing.

Techniques

Set up the equation/formula and compute; no special trick needed