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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.

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    $\begingroup$ The line $y-x=0$? $\endgroup$ Commented Mar 27, 2019 at 10:20
  • $\begingroup$ Oh yes! I'll edit that. $\endgroup$
    – idunno
    Commented Mar 27, 2019 at 10:22

1 Answer 1

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You have made an error in calculating the determinant.

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

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