Question

Se dau punctele A(2,-3) B (m +1,2m), C (1,5) : a)Sa se determine ecuatia dreptei AC ,b) Pentru ce valori ale parametrului m punctele A, B, C sunt coliniare. c) sa se determine triunghiul ABC cu aria 22,5.

Answer 1

[tex]a)\:AC:\frac{x-x_C}{x_A-x_c}=\frac{y-y_C}{y_A-y_C}\\ \\ \frac{x-1}{2-1}=\frac{y-5}{-3-5}\\ \\ -8(x-1)=y-5\\ -8x+8=y-5\\ AC:y+8x-16=0\\ b)A,\:B,\:C\:coliniare\Leftrightarrow\det(Delta)=0\\ \\ \Delta= \left[\begin{array}{ccc}x_A&y_A&1\\x_B&y_B&1\\x_C&y_C&1\end{array}\right] = \left[\begin{array}{ccc}2&-3&31\\m+1&2m&1\\1&5&1\end{array}\right] \\ \\ det(\Delta)=4m+5(m+1)-3-(2m+10-3m-3)=\\ =4m+5m+5-3-2m-10+3m+3=10m-5=0\Leftrightarrow \\ 5(2m-1)=0\\ \Leftrightarrow 2m=1\Leftrightarrow m=\frac{1}{2} [/tex]
[tex]c)\mathcal{A}=\frac{1}{2}det(\Delta)=22,5=\frac{225}{10}=\frac{45}{2}\\ \frac{1}{2} det(\Delta)=\frac{45}{2}\\ \\ det(\Delta)=45\\ \\ 10m-5=45\\ 10m=50\\ m=5[/tex]