Задача 62489 find the slope of the line belowing to...

61ccafbaf9ee7a38ce6b06d4

29.12.2021 21:59:20

find the slope of the line belowing to the two bunches (2x-2)x+(5k-4)y-3k-6=0 and (k+5)x+(7k-5)y+3k+10=0

5f3ea7e3faf909182968ddd9

30.12.2021 18:01:37

★

(2x–2)x+(5k–4)y–3k–6=0 ⇒ (5k-4)y=-(2k-2)x+(3k-6)

y=-(2k-2)x/(5k-4) + (3k-6)/(5k-4)

Угловой коэффициент

k_(1)=-(2k-2)/(5k-4)



(k+5)x+(7k–5)y+3k+10=0 ⇒ (7k–5)y=- (k+5)x-(3k+10)

y=-(k+5)x/(7k-5) - (3k+10)/(7k-5)

Угловой коэффициент

k_(2)=-(k+5)/(7k-5)


Прямые параллельны, если

k_(1)=k_(2)

-(2k-2)/(5k-4)=-(k+5)/(7k-5) ⇒

(2k-2)*(7k-5)=(5k-4)*(k+5)

14k^2-14k-10k+10=5k^2-4k+25k-20

9k^2-45k+30=0

3k^2-15k+10=0

D=225+4*3*10=105

k=[b](15 ± sqrt(105)/6[/b]