Question

4. Find the equations of the sides of the triangle whose vertices are (-1,8),(4.-2) and (-5.-3).​

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Considering a Triangle whose vertices are :-

P(-1 , 8)

Q(4 , -2)

R(-5 , -3)

Here we get two equations which help us to solve furthur steps

${\boxed{\sf\:{Equation\:of\;PQ=\dfrac{y-8}{x+1}}}}$

${\boxed{\sf\:{Equation\:of\;PQ=\dfrac{-2-8}{4+1}}}}$

5(y - 8) + 10(x + 1) = 0

10x + 5y - 30 = 0

2x + y - 6 = 0

${\boxed{\sf\:{Equation\:of\;QR=\dfrac{y+2}{x-4}}}}$

${\boxed{\sf\:{Equation\:of\;QR=\dfrac{-3+2}{-5-4}}}}$

-9(y + 2) = -(x - 4)

x - 9y - 22 = 0

${\boxed{\sf\:{Equation\:of\;PR=\dfrac{y-8}{x+1}}}}$

${\boxed{\sf\:{Equation\:of\;PR=\dfrac{-3-8}{-5+1}}}}$

-4(y - 8) = -11(x + 1)

11x - 4y + 43 = 0

Answer 2

Answer:

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