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(A) The area of $	riangle ABC$ is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 8 	imes 1 = 4$. The line $x=a$

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$3$

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(A) The area of $	riangle ABC$ is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 8 	imes 1 = 4$. The line $x=a$ intersects $BC$ at $D(a, 1)$ and $AC$ at $E(a, a/9)$. The triangle formed to the right of the line $x=a$ is $	riangle DEC$ with vertices $D(a, 1)$,$E(a, a/9)$,and $C(9, 1)$. The area of $	riangle DEC$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (9-a) 	imes (1 - a/9) = \frac{1}{2} 	imes (9-a) 	imes \frac{9-a}{9} = \frac{(9-a)^2}{18}$. Since the line $x=a$ divides the triangle into two equal areas,the area of $	riangle DEC$ must be half the total area: $\frac{(9-a)^2}{18} = \frac{1}{2} 	imes 4 = 2$. $(9-a)^2 = 36$. Taking the square root,$9-a = 6$ (since $a < 9$),so $a = 3$.

Consider a $\triangle ABC$ in the $XY$-plane with vertices $A=(0,0)$,$B=(1,1)$,and $C=(9,1)$. If the line $x=a$ divides the triangle into two parts of equal area,then $a$ equals

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