1957 AMC 12 Problem 25

Question

The vertices of triangle

PQR

have coordinates as follows:

P(0,a),

Q(b,0),

R(c,d),

where

a,

b,

c,

and

d

are positive. The origin and point

R

lie on opposite sides of

PQ.

The area of triangle

PQR

may be found from the expression:

\dfrac{ab+ac+bc+cd}{2}

\dfrac{ac+bd-ab}{2}

\dfrac{ab-ac-bd}{2}

\dfrac{ac+bd+ab}{2}

\dfrac{ac+bd-ab-cd}{2}

Use the determinant formula for the area of a triangle from its coordinates

The opposite-side condition determines the sign inside the absolute value

The signed doubled area is

\begin{vmatrix} 0&a&1\\ b&0&1\\ c&d&1 \end{vmatrix} =ab-ac-bd.

The line

PQ

has equation

ax+by=ab.

Since the origin gives a value below

ab

and

R

is on the opposite side,

ac+bd>ab.

Thus the area is

\frac{ac+bd-ab}{2}.

Therefore, the correct answer is B.

Problem 25 in other years