Question

Графік функції y=kx+l проходить через точки (3; -4) і (-12; -9)
Можно быстренько?

Answer 1

$y=kx+l\; \; ,\; \; A(3;-4)\; \; ,\; \; B(-12;-9)\\\\\\\left\{\begin{array}{l}-4=3k+l\\-9=-12k+l\end{array}\right\; \; \left\{\begin{array}{l}-16=12k+4l\\-9=-12k+l\end{array}\right\; \oplus\; \left\{\begin{array}{l}-25=5l\\3k=-4-l\end{array}\right\\\\\\\left\{\begin{array}{l}l=-5\\3k=1\end{array}\right\; \; \left\{\begin{array}{l}l=-5\\k=\frac{1}{3}\end{array}\right\\\\\\\boxed {\; y=\frac{1}{3}\, x-5\; }$