Question

dado os pontos A(k-3, k) B(-3,5), situado na reta "r" e que possui o coeficiente angular igual a -2/3. determine as coordenadas do ponto A.

Answer 1

Construindo a equação da reta:

$y-y_{0} = m \cdot (x-x_{0}) \\\\ y-5 = -\dfrac{2}{3} \cdot (x+3) \\\\\\ y-5 = -\dfrac{2x}{3}-\dfrac{2 \cdot 3}{3} \\\\\\ y-5 = -\dfrac{2x}{3}-2 \\\\ y = -\dfrac{2x}{3}-2+5 \\\\ \boxed{y = -\dfrac{2x}{3}+3}$

Possuindo a equação da reta, conseguimos jogar o ponto A.

$y = -\dfrac{2x}{3}+3 \\\\\\ k = -\dfrac{2 \cdot (k-3)}{3}+3 \\\\\\ k = \dfrac{-2k+6}{3}+\dfrac{9}{3} \\\\\\ 3k = -2k+15 \\\\ 5k = 15 \\\\ \boxed{k = 3 }$

Então o ponto A é:

$A(k-3, \ k) \\\\ A(3-3, \ 3) \\\\ \boxed{\boxed{A(0,3)}}$

Answer 2

Sabemos que pela definição de coeficiente angular, o coeficiente angular $m$ de uma reta que passa pelos pontos $A$ e $B$ é dado por

$m=\dfrac{\Delta y}{\Delta y}=\dfrac{y_{_B}-y_{_A}}{x_{_B}-x_{_A}}~~~~~~(\text{com } x_{_B}\ne x_{_A})$


Para esta questão, devemos ter

$m=-\,\dfrac{2}{3}\\\\\\ \dfrac{\Delta y}{\Delta y}=-\,\dfrac{2}{3}\\\\\\ \dfrac{y_{_B}-y_{_A}}{x_{_B}-x_{_A}}=-\dfrac{2}{3}\\\\\\ \dfrac{5-k}{-3-(k-3)}=-\dfrac{2}{3}\\\\\\ \dfrac{5-k}{-\diagup\!\!\!\! 3-k+\diagup\!\!\!\! 3}=-\dfrac{2}{3}\\\\\\\dfrac{5-k}{-k}=-\dfrac{2}{3}\\\\\\3(5-k)=-2\cdot (-k)\\\\ 15-3k=2k\\\\ -3k-2k=-15\\\\ -5k=-15\\\\ k=\dfrac{-15}{-3}\\\\\\ \therefore~~\boxed{\begin{array}{c}k=3 \end{array}}$


Portanto, as coordenadas do ponto $A$ são

• k - 3 = 3 - 3 = 0

• k = 3


$A(0,\,3)$


Bons estudos! :-)