Решите любое задание, любому буду рад

Решите задачу:

$5.1)\; \; A(-4,-3)\; ,\; B(-1,5)\; ,\; C(6,-2)\\\\a)\; \; AB:\; \; \frac{x+4}{-1+4}=\frac{y+3}{5+3} \; \; ,\; \; \; \frac{x+4}{3} =\frac{y+3}{8} \\\\b)\; \; AH\perp BC\; \; \to \; \; \overline {BC}=\vec{n} =(7,-7)\; \; ili\; \; \lambda =\frac{1}{7}\; \; \to \; \; \vec{n}=(1,-1)\\\\AH:\; \; 1\cdot (x+4)-1\cdot (y+3)=0\; \; \to \; \; \underline {x-y+1=0}\\\\c)\; \; x_{E}= \frac{x_{A}+x_{C}}{2}=\frac{-4+6}{2}=1 \; ,\\\\y_{E}= \frac{y_{A}+y_{C}}{2}=\frac{-3-2}{2}=-2,5\\\\BE:\; \; \frac{x+1}{1+1}=\frac{y-5}{-2,5-5}\; \; ,\; \; \frac{x+1}{2}=\frac{y-5}{-7,5}$

$BE:\; \; \frac{x+1}{4} =\frac{y-5}{15}\\\\15(x+1)=4(y-5)\\\\BE:\; \; \underline {15x-4y+35=0}\\\\d)\; \; CN\parallel AB\; \; \to \; \; \overline {AB}=\vec{s}_{CN}=(3,8)\; ,\; \; C(6,-2)\in CN\\\\CN:\; \; \frac{x-6}{3} = \frac{y+2}{8}\\\\8(x-6)=3(y+2)\; \; , \; \; CN:\; \; \underline {8x-3y-54=0}$

$e)\; \; N=BE\cap CN\; \; \to \; \; \; \left \{ {{15x-4y=-35|\cdot 3 } \atop {8x-3y=54|\cdot (-4)}} \right. \; \oplus \; \left \{ {{13x=-321} \atop {3y=8x-54}} \right. \\\\ \left \{ {{x=-\frac{321}{13}} \atop {y=-\frac{1090}{13}}} \right. \; \; ,\; \; N\Big (-\frac{321}{13},-\frac{1090}{13}\Big )$