Two vertices of a triangle are $(3,-2)$ and $(-2,3)$ and its orthocentre is $(-6,1)$. Then the third vertex cannot lie on

Question

A) $4x+y=2$

B) $5x+y=2$

C) $3x+y=3$

D) $6x+y=0$

I found the slope of the line joined by the given vertices to be $-1$

The line joining the third vertex and the orthocenter will have slope $1$

It’s equation will be $$y-1=x+6$$ $$x-y+7$$

The third vertex must lie on this line. This, however doesn’t give much information on which line the vertex CANNOT lie. One would assume the line parallel to this line should be the right answer, but there are none in the options, and I have no idea on what needs to be done

Answer 1

Note that the third vertex would be the intersection of the equation $y=x+7$ you derived and the equations given in the answer choices. The equations (A), (C) and (D) all give the same intersection $(-1,6)$ with $y=x+7$. It is straightforward to verify that the two other altitude lines, with the slopes $\frac12$ and $-\frac13$, are perpendicular to their respective sides of the slope $-2$ and $3$. Therefore, the third vertex is on (A), (C) and (D).

For the equation $5x+y=2$, the intersection is $(-\frac56, -\frac{37}6)$. Unlike the other three equations, the resulting altitudes are not perpendicular to the corresponding sides. Thus, the third vertex can not lie on (B).

Answer 2

Let C be the third vertex. [Its exact position is not known, but I just happen to have the tool to locate it. Drawing a moderately accurte picture can make the explanation a bit easier.]

Through all the vertices, respective altitudes are drawn (the red dotted lines). They will go through the ortho-center H.

We obtain 2 circles. The green circle with AH as diameter has the equation (x + 6)(x – 3) +(y – 1)(y +2) = 0. The blue circle with AB as diameter has the equation …..

Solving the two, we get the equation (#1) of A(B’)C.

By point slope form, the equation (#2) of H(X)C can also be obtained.

Co-ordinates of C = (p, q) can therefore be found after solving (#1) and (#2).

It remains to test which option is satisfied by (p, q).

---

Remarks:- Using a standard approach (like the one above) is quite involoved alraady. Added to the diffiuclty is the line representing the correct option is totally unrelated to any lines (either given or constructed) in the sense that it is neither perpenicular nor parallel nor bisected nor....

It is entirely worthless to spend the time and effort on this for just an 1 MC mark return. Maybe the question setter has some special technique that can answer this one in a minute or two.