0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ The line $y-x=0$? $\endgroup$ – Shubham Johri Mar 27 at 10:20
  • $\begingroup$ Oh yes! I'll edit that. $\endgroup$ – JC2000 Mar 27 at 10:22
1
$\begingroup$

You have made an error in calculating the determinant.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.