# Reconciling two solutions to locus problem involving area of triangle and equations of the sides.

I was solving the following problem where I was unable to reach the same conclusion using two methods.

The problem :

The area of a triangle formed by the intersection of the line parallel to X-axis and passing through $$P(h,k)$$ with the lines $$y-x=0$$ and $$x+y=2$$ is $$4h^2$$. Find the locus of point $$P$$.

My solution :

I found the points of intersection to be $$A(1,1),B(k,k),C(2-k,k)$$ and then I used the formula given below to equate the area to $$4h^2$$. On simplification I got the following relationship as the locus : $$8h^2 = (2k-1)(k-1)$$

The formula :

\begin{align*} \text{Area} &= \frac12 \big| (x_A - x_C) (y_B - y_A) - (x_A - x_B) (y_C - y_A) \big|\\ &= \frac12 \big| x_A y_B + x_B y_C + x_C y_A - x_A y_C - x_C y_B - x_B y_A \big|\\ &= \frac12 \big|\det \begin{bmatrix} x_A & x_B & x_C \\ y_A & y_B & y_C \\ 1 & 1 & 1 \end{bmatrix}\big| \end{align*}

The solution in my book :

Since $$y+x=2$$ and $$y=x$$ are perpendicular, the area of the triangle can also be found using $$(1/2)AB*AC$$ where $$AB = \sqrt{2}*|k-1|$$ and $$AC=\sqrt{2}|k-1|$$.

This on simplification gives the locus to be $$2x = +-(y-1)$$.

My questions :

1. Why does my answer vary from the given solution? Is there something I have overlooked in my solution that causes this?

2. If my answer is the same as the given solution then how do I write it as such.

• The line $y-x=0$? – Shubham Johri Mar 27 at 10:20
• Oh yes! I'll edit that. – JC2000 Mar 27 at 10:22

$$\frac12 \big|\det \begin{bmatrix} x_A & x_B & x_C \\ y_A & y_B & y_C \\ 1 & 1 & 1 \end{bmatrix}\big|=\frac12 \big|\det \begin{bmatrix} 1&k&2-k\\ 1&k&k\\ 1&1&1 \end{bmatrix}\big|=(k-1)^2$$