Question

ABC is a variable triangle such that A is (1, 2), B and C lie on line y = x + $\lambda$ (where $\lambda$ is a variable), then locus of the orthocenter of triangle ABC is

• A. $(x - 1)^2 + y^2 = 4$
• B. $x + y = 3$
• C. $2x - y = 0$
• D. none of these

Answer 1

Answer: (B)

x + y = 3

Answer 2

The problem asks for the locus of the orthocenter of a variable triangle ABC, where vertex A is fixed at (1, 2) and vertices B and C lie on the line $y = x + \lambda$, where $\lambda$ is a variable.

Let H be the orthocenter of triangle ABC. The orthocenter is the intersection point of the altitudes of the triangle.

1. Determine the slope of the line BC: The equation of the line on which B and C lie is $y = x + \lambda$. This can be rewritten as $x - y + \lambda = 0$. The slope of the line BC, $m_{BC}$, is 1.

2. Determine the equation of the altitude from A to BC: The altitude from vertex A to side BC (let's call it AD) is perpendicular to BC. The slope of the altitude AD, $m_{AD}$, will be the negative reciprocal of the slope of BC. $m_{AD} = -\frac{1}{m_{BC}} = -\frac{1}{1} = -1$. The altitude AD passes through the fixed point A(1, 2) and has a fixed slope of -1. Using the point-slope form of a line ($y - y_1 = m(x - x_1)$), the equation of the altitude AD is: $y - 2 = -1(x - 1)$ $y - 2 = -x + 1$ $x + y = 3$

3. Find the locus of the orthocenter: The orthocenter H of triangle ABC must lie on all three altitudes. Since the altitude from A (AD) is a fixed line with the equation $x + y = 3$, the orthocenter H must always lie on this line. As $\lambda$ varies, the positions of B and C change, and thus the triangle ABC changes. However, the line BC always maintains a slope of 1. Consequently, the altitude from A always remains the same fixed line $x + y = 3$. Therefore, the locus of the orthocenter of triangle ABC is the line $x + y = 3$.

Comparing this with the given options, the derived locus matches option (B).

The final answer is $x + y = 3$.

Explanation of the solution:

The line BC has a constant slope of 1, irrespective of $\lambda$. The altitude from vertex A(1, 2) to BC must be perpendicular to BC, so its slope is -1. Using the point-slope form, the equation of this altitude is $y - 2 = -1(x - 1)$, which simplifies to $x + y = 3$. Since the orthocenter must lie on every altitude, it must lie on this fixed altitude. Hence, the locus of the orthocenter is $x + y = 3$.