Question

Let triangle $ABC (\angle C = \frac{\pi}{2})$ is right angled triangle of area equal to 12 units lie in xy plane so that its perpendicular sides are parallel to coordinates axes and the medians through A and B lie on the lines $y = 3x + 1$ and $y = mx + 2 (m \in I)$ respectively. If the coordinates of vertex $C$ is $(a, b)$ then $\frac{1}{|ab|}$ is equal to

Answer 1

27/50

Answer 2

Let the vertices be $A(x_A, y_A)$, $B(x_B, y_B)$, and $C(a, b)$. Since $\angle C = \frac{\pi}{2}$ and the perpendicular sides are parallel to the coordinate axes, the sides $AC$ and $BC$ are parallel to the coordinate axes. Then we have two cases: Case 1: $A=(a, y_A), B=(x_B, b), C=(a,b)$ Case 2: $A=(x_A, b), B=(a, y_B), C=(a,b)$

We are given that the median through A lies on the line $y = 3x + 1$ and the median through B lies on the line $y = mx + 2$. Let's analyze Case 2: $A=(x_A, b), B=(a, y_B), C=(a,b)$. The midpoint of BC is $M_A(a, \frac{y_B+b}{2})$. The median through A connects $A(x_A, b)$ to $M_A(a, \frac{y_B+b}{2})$. Since this median lies on the line $y=3x+1$, we have $b = 3x_A+1$ and $\frac{y_B+b}{2} = 3a+1$. The midpoint of AC is $M_B(\frac{x_A+a}{2}, b)$. The median through B connects $B(a, y_B)$ to $M_B(\frac{x_A+a}{2}, b)$. Since this median lies on the line $y=mx+2$, we have $y_B = ma+2$ and $b = m(\frac{x_A+a}{2})+2$.

From $b=3x_A+1$, we get $x_A=\frac{b-1}{3}$. From $\frac{y_B+b}{2} = 3a+1$, we get $y_B+b = 6a+2$. From $y_B = ma+2$, we get $ma+2+b = 6a+2$, so $b = a(6-m)$. From $b = m(\frac{x_A+a}{2})+2$, we get $2b = mx_A+ma+4$, so $mx_A = 2b-ma-4$. Substituting $x_A = \frac{b-1}{3}$, we have $m(\frac{b-1}{3}) = 2b-ma-4$, so $m(b-1) = 6b-3ma-12$. Then $mb-m = 6b-3ma-12$, so $b(m-6) = -3ma-12+m$. Substituting $b = a(6-m)$, we have $a(6-m)(m-6) = -3ma-12+m$, so $-a(m-6)^2 = -3ma-12+m$. $-a(m^2-12m+36) = -3ma-12+m$, so $-am^2+12am-36a = -3ma-12+m$. $-am^2+15am-36a+12-m = 0$, so $a(-m^2+15m-36) = m-12$, so $a(-(m-3)(m-12)) = m-12$. If $m \neq 12$, then $a(-(m-3)) = 1$, so $a = \frac{-1}{m-3} = \frac{1}{3-m}$. Then $b = a(6-m) = \frac{6-m}{3-m}$.

Since area = 12, $\frac{1}{2}|x_A-a||y_B-b|=12$, so $|x_A-a||y_B-b|=24$. $x_A = \frac{b-1}{3} = \frac{\frac{6-m}{3-m}-1}{3} = \frac{\frac{6-m-3+m}{3-m}}{3} = \frac{3}{3(3-m)} = \frac{1}{3-m}$. $x_A - a = \frac{1}{3-m} - \frac{1}{3-m} = 0$. This implies $x_A=a$, which is not possible.

If $m = 12$, then $0 = m-12 = 0$, so $m=12$. $b=3x_A+1$. $y_B = 12a+2$. $\frac{y_B+b}{2} = 3a+1 \implies y_B+b = 6a+2$. $b = 12(\frac{x_A+a}{2})+2 \implies b = 6x_A+6a+2$. $b-2=6x_A+6a$. $y_B+b = 6a+2$, so $y_B = 6a+2-b$. $b = 3x_A+1$, so $x_A = \frac{b-1}{3}$. $2b = 12(\frac{x_A+a}{2})+4$, so $2b = 6x_A+6a+4$, so $b-2=3x_A+3a$. $x_A = \frac{b-1}{3}$, so $b-2 = b-1+3a$, so $3a=-1$, so $a=-1/3$. Then $y_B = 12a+2 = 12(-1/3)+2 = -4+2 = -2$. $b = a(6-m) = (-1/3)(6-12) = (-1/3)(-6)=2$. Area = $\frac{1}{2}|x_A-a||y_B-b| = \frac{1}{2}|\frac{b-1}{3}-a||ma+2-b| = 12$. If $a=-1/3$ and $b=2$, then $x_A = \frac{2-1}{3} = \frac{1}{3}$. $y_B = 12a+2 = 12(-1/3)+2 = -4+2 = -2$. Area = $\frac{1}{2}|1/3-(-1/3)| |-2-2| = \frac{1}{2}|2/3||-4| = \frac{1}{2}(2/3)(4) = 4/3$. This contradicts the area being 12.

Let's consider $A=(x,b)$, $B=(a,y)$. Area is $|x-a||y-b|=24$. $M_A=(a, \frac{y+b}{2})$. $M_B=(\frac{x+a}{2}, b)$. $A$ on $y=3x+1$: $b=3x+1$. $B$ on $y=mx+2$: $y=ma+2$. $M_A$ on $y=3x+1$: $\frac{y+b}{2} = 3a+1$. $M_B$ on $y=mx+2$: $b=m(\frac{x+a}{2})+2$. $b=3x+1$, so $x=(b-1)/3$. $y=ma+2$. $\frac{ma+2+b}{2}=3a+1$, so $ma+2+b=6a+2$, so $b=6a-ma = a(6-m)$. $b=m(\frac{x+a}{2})+2$, so $2b=mx+ma+4$, so $2b = m\frac{b-1}{3}+ma+4$, so $6b = mb-m+3ma+12$. Then $b(6-m) = 3ma+12-m$. $a(6-m)^2 = 3ma+12-m$. $a(36-12m+m^2) = 3ma+12-m$, so $36a-12ma+am^2=3ma+12-m$, so $am^2-15ma+36a+m-12=0$. $a(m^2-15m+36) = -m+12$, so $a(m-3)(m-12) = -(m-12)$. If $m=12$, this is always true. If $m \neq 12$, $a(m-3) = -1$, so $a=\frac{-1}{m-3} = \frac{1}{3-m}$. Then $b = a(6-m) = \frac{6-m}{3-m}$. $|x-a||y-b| = 24$, so $|\frac{b-1}{3}-a||ma+2-b| = 24$. $x-a=\frac{b-1}{3}-a = \frac{\frac{6-m}{3-m}-1}{3}-\frac{1}{3-m} = \frac{3}{3(3-m)}-\frac{1}{3-m} = 0$. So impossible.

Let $x=a+2, y=b-12$. $xy=24$. Then $b=3(a+2)+1=3a+7$. Median B connects $B(a,b-12)$ to $(\frac{a+2+a}{2}, b) = (a+1, b)$. Slope $\frac{12}{1}=12$, so $m=12$. Then $y=12x+2$. $b=12(a+1)+2=12a+14$. $3a+7=12a+14$, so $9a=-7$, so $a=-7/9$. Then $b=14/3$. $|ab|=98/27$, so $1/|ab|=27/98$. If $x=a-2, y=b+12$. $b=3(a-2)+1=3a-5$. $b+12=12a+2$, so $b=12a-10$. $3a-5=12a-10$, so $9a=5$, so $a=5/9$. $b=-10/3$. $|ab|=50/27$, so $1/|ab|=27/50$.

Final Answer: The final answer is $\boxed{\frac{27}{50}}$