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I have a triangle in $ xy $ plane, whose vertices being

$$ A(k, –3k)$$ $$ B(5, k) $$ $$ C(–k, 2) $$

Where $k$ is an integer.

And area is $28$ sq. units.

How to find the coordinates of its orthocentre?

What I've done

By using area formula, I got $k=2$

Fghj
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2 Answers2

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Hint

Since the area is given as well as the coordinates, this can help you determine the value of $k$ from the following: $$\frac{1}{2}\begin{vmatrix}k&-3k&1\\5&k&1\\-k&2&1\end{vmatrix}=28.$$

Once you have the $k$, it is a matter of determining the intersection of two altitudes of the triangle.

For example, to get the altitude from $C$ onto $AB$, you want to find a line that passes through $C$ and is perpendicular to $AB$. Slope of $AB$ is $m_{AB}=\frac{4k}{5-k}$. So your line must of the form $$y-2=-\frac{5-k}{4k}(x+k).$$ Similarly you can get the other altitude (say from $B$ onto $AC$) and then find the intersection of these two altitudes. By definition, the intersection is the orthocenter.

Anurag A
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  • "....You want to find a line that passes through C and is perpendicular to AB. Slope of AB....." Why to find slope? – Fghj Aug 11 '17 at 19:03
  • @RaviPrakash to determine a line, you need two (independent) pieces of information. So if you want the altitude from $C$, you know that your line passes through $C$, so what's the other piece of information you need..? – Anurag A Aug 11 '17 at 19:05
  • I think it's the side in which it will fall – Fghj Aug 11 '17 at 19:07
  • @RaviPrakash well, if it is the altitude from $C$ then it will fall on $AB$. So you know the side but what you don't know is where it will fall on $AB$ so you don't have the point on $AB$ through which this altitude will pass. So you still need some other information!! – Anurag A Aug 11 '17 at 19:08
  • I think the 'angle' or 'inclination'?? – Fghj Aug 11 '17 at 19:10
  • @RaviPrakash I would recommend you to draw a picture and think about what is an "orthocenter". It will help you figure it out, – Anurag A Aug 11 '17 at 19:10
  • I can make one in my mind, it's a perpendicular line originating from the vertex of ∆ to opposite side – Fghj Aug 11 '17 at 19:12
  • Anurag, please figure out the information about which you were talking – Fghj Aug 11 '17 at 19:14
  • After reading the chapter Straight Lines and Circles, I got what you were saying and its absolutely correct :) – Fghj Dec 08 '17 at 17:33
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Once you get $k=2$ from the shoelace formula, you also get that the $BC$ side is horizontal, hence the abscissa of the orthocenter is the abscissa of $A$, i.e. $2$. Since the slope of the $CA$ side is $-2$, the orthocenter $H$ lies on the line through $B$ with slope $\frac{1}{2}$. From $y=\frac{1}{2}(x-5)+2$ you get $H=\left(2,\frac{1}{2}\right)$. Anyway there is another solution associated with $k=-\frac{23}{5}$.

Jack D'Aurizio
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