I came across a question today.

Two mutually perpendicular straight lines through the origin forms an isosceles triangle with the line $2x + y = 5$. Then the area of the triangle is ?

I know that it can be solved by finding the length ($l$) of perpendicular from origin to the line and then the length of the hypotenuse and then using $\text{Area }= \frac{ab}{2} = l^2$. And the area came out to be $5$ (which is a correct answer).

But I tried a different approach. I took the perpendicular straight lines as $x$-axis and $y$-axis and found the coordinates of the line on the axes using intercept form of line. They came out to be $(\frac{5}{2}, 0)$ and $(0,5)$.

Now as these points works as base and altitude of the triangle. So $\text{Area}=\left(\frac{1}{2}\right)\left(\frac{5}{2}\right)\left(5\right)=\frac{25}{4}$

What is the mistake in the second solution?

  • 4
    $\begingroup$ The x-axis, y-axis and $2x+y=5$ do not form a isosceles triangle $\endgroup$ – imranfat Feb 21 '16 at 21:11
  • $\begingroup$ @imranfat thanks... $\endgroup$ – manshu Feb 21 '16 at 21:13

The line $2x+y=5$ can be written as $\frac{y}{5} +\frac{x}{5/2}=1$ or $y=-2x+5$

The slope is $m1 = -2$

A normal to this line has slope $m2 = 1/2$

This means that for the other two lines or sides of the triangle, the slopes are:

$m3 = tan(\text{atan}(m1)+\pi/4)$

$m4 = tan(\text{atan}(m1)-\pi/4)$

In degrees:

$m1 = -63.4349^\circ$

$m2 = +26.5650^\circ$

$m3 = +71.5650^\circ$

$m4 = -18.4349^\circ$

Now we are in conditions to calculate intersection points, lenghts, and area.

  • 1
    $\begingroup$ You answered a different question. Manshu answered the question asked, which was much easier. But your answer was nice, so I'll upvote it. $\endgroup$ – marty cohen Jan 26 '17 at 0:49

Thanks to @imranfat for answering in the comments.

It is because the $x$-axis (given by $y=0$), $y$-axis (given by $x=0$), and $2x+y=5$ do not form a isosceles triangle when it is clearly given in the question that the triangle should be isosceles.


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