Answer :
[tex]The\ slope-point\ form:y-y_1=m(x-x_1)\\-----------------------------\\6A:\\m=-5;\\(-2;\ 6)\to x_1=-2;\ y_1=6\\\\y-6=-5[x-(-2)]\\y-6=-5(x+2)\\y-6=-5x-5(2)\\y-6=-5x-10\ \ \ |add\ 6\ to\ both\ sides\\\boxed{y=-5x-4}\\\\put\ the\ coordinates\ of\ the\ point\ (r;-4)\to x=r\ and\ y=-4:\\-4=-5r-4\ \ \ |add\ 4\ to\ both\ sides\\0=-5r\\-5r=0\ \ \ \ |divide\ both\ sides\ by\ (-5)\\\boxed{r=0}\leftarrow answer[/tex]
[tex]6B.\\ The\ same\ method:\\y-(-8)=-8(x-5)\\y+8=-8(x)-8(-5)\\y+8=-8x+40\ \ \ \ |subtract\ 8 from\ both\ sides\\\boxed{y=-8x+32}\\\\(r;-6)\to x=r;\ y=-6\\\\-6=-8(r)+32\\-6=-8r+32\ \ \ \ |subrtact\ 32\ from\ both\ sides\\-38=-8r\\8r=38\ \ \ \ \ |divide\ both\ sides\ by\ 8\\\boxed{r=4.75}[/tex]
[tex]6B.\\ The\ same\ method:\\y-(-8)=-8(x-5)\\y+8=-8(x)-8(-5)\\y+8=-8x+40\ \ \ \ |subtract\ 8 from\ both\ sides\\\boxed{y=-8x+32}\\\\(r;-6)\to x=r;\ y=-6\\\\-6=-8(r)+32\\-6=-8r+32\ \ \ \ |subrtact\ 32\ from\ both\ sides\\-38=-8r\\8r=38\ \ \ \ \ |divide\ both\ sides\ by\ 8\\\boxed{r=4.75}[/tex]
