Which of the following point is outside the triangle?

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If $P(6,7),Q(2,3)\ \text{and}\ R(4,-2)$ be the vertices of the triangle , then which of the point is not contained in the triangle?

$a.)(4,3)\quad \quad \quad \quad b.)(3,3)\\ c.)(4,2)\quad \quad \quad \quad \color{green}{d.)(6,1)}\\$

How can i find out with pen and paper without using graph paper.

I look for a simple method , i have studied maths up to $12th$ grade.

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One approach is to calculate the equations of the lines connecting each of the vertices of the triangle. In your case you would calculate the equations of the lines: $PQ, PR$ and $QR$.

You could then substitute the $x$ coordinate of the point in question into each of these lines and compare this value with the $y$ coordinate of the point in question. This will help you determine if the given point lies above, on, or below each of the lines.

This should give you enough information to determine if the point lies inside or outside the triangle.


NOTE: This is a lot more work then simply graphing the points as suggested by @TBrendle above

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Without drawing the points we can use an operator $\operatorname{pqr}(P,Q,R)\to (Q_x-P_x)(R_y-P_y)-(Q_y-P_y)(R_x-P_x)$ which sign indicates counterclockwise (if positive) or clockwise (if negative) order of points $P,Q,R$.

If points $P,Q,R$ are arranged in the counterclockwise order, then $\operatorname{pqr}(P,Q,R)>0$ and then for the point $T$ all three tests $\operatorname{pqr}(P,Q,T)$,$\operatorname{pqr}(Q,R,T)$, and $\operatorname{pqr}(R,P,T)$ are positive (while moving along $P-Q-R-P$ path, the point $T$ is always on the left). Any negative result means the point $T$ is on the right, that is, outside the $\triangle PQR$.

For the given example, all tests are positive exept the only one test for the point $D$: $\operatorname{pqr}(R,P,D)=(6-4)(1+2)-(7+2)(6-4)=-12$, hence only the point $D$ is outside of $\triangle PQR$.

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See also shoelace formula.