Answer:
The lines meet at (-1, -3)
Step-by-step explanation:
Line A :
[tex](x_1, y_1) = (3, 3) \ ; \ slope, \ m_A = \frac{3}{2}\\\\Equation \ of \ line \ A : (y - 3) = \frac{3}{2}(x - 3)[/tex]
[tex]2(y - 3) = 3(x-3)\\2y - 6 = 3x - 9\\2y = 3x - 9 + 6\\2y = 3x -3[/tex]
Line B :
[tex](x_2,y_2) = (-4, -2) \ ; \ slope , m_B = -\frac{1}{3}\\\\Equation \ of\ line \ B: (y -(-2)) = -\frac{1}{3}(x -(-4))[/tex]
[tex]3(y + 2) = -1(x+4)\\3y + 6 = -x -4\\3y = -x - 4 - 6 \\3y = -x - 10[/tex]
Solve for x and y from the linear equation to find where line A and line B meets :
2y = 3x - 3 => 3x - 2y = 3 ------- (1)
3y = -x - 10 => -x = 3y + 10
=> x = - 3y - 10 --------(2)
Substitute (2) in (1) : => 3(- 3y - 10) - 2y = 3
-9y - 30 -2y = 3
-11y = 3 + 30
-11y = 33
y = -3
Substitute y in (2) : => x = -3 (-3) - 10 = 9 - 10 = -1
