Question

find the coordinates of the vertices of a triangle the equation of whose sides are x+y-4=0,2x-y+3=0andx-3y+2=0​

Answer 1

Answer:

A( - 1.4 , 0.2 ) ; B( $\frac{1}{3}$ , $\frac{11}{3}$ ) ; C( 2.5 , 1.5 )

Step-by-step explanation:

ΔABC

1). A( $x_{A}$, $y_{A}$ )

x - 3y = - 2 ........ (1)

2x - y = - 3 ........ (2)

(1) × 2 - (2)

- 5y = - 1 ⇒ $y_{A}$ = $\frac{1}{5}$ = 0.2

2x - $\frac{1}{5}$ = - 3 ⇒ $x_{A}$ = - 1.4

A( - 1.4 , 0.2 )

2). B( $x_{B}$ , $y_{B}$ )

2x - y = - 3 ......... (3)

x + y = 4 ............. (4)

(3) + (4)

3x = 1 ⇒ $x_{B}$ = $\frac{1}{3}$

$\frac{1}{3}$ + y = 4 ⇒ $y_{B}$ = $\frac{11}{3}$

B( $\frac{1}{3}$ , $\frac{11}{3}$ )

3). C( $x_{C}$ , $y_{C}$ )

x + y = 4 ............. (5)

x - 3y = - 2 ........ (6)

(5) - (6)

4y = 6 ⇒ $y_{C}$ = 1.5

x + 1.5 = 4 ⇒ $x_{C}$ = 2.5

C( 2.5 , 1.5 )