Point of Intersection Formula

Answer: The two line equations are given as 

2x+4y+6=0 and 2x+3y+4=0

Comparing these two equations with 

a₁x+b₁y+c₁and a₂x+b₂y+c₂, We get 

a₁= 2, b₁= 4, c₁= 6, a₂ = 2, b₂ =3, c₂ = 4

Point of intersection formula is given as

(x, y) = ((b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁), (c₁a₂ – c₂a₁) / (a₁b₂ – a₂b₁))

Substituting the values of a₁, b₁, c₁, a₂, b₂, c₂ in the intersection formula

(x, y) = ((4×4 -36) / (23 – 24), (62 – 42) / (23 – 2 4))

(x, y) = ((16-18) / (6-8), (12-8) / 6-8))

(x, y) = (-2/-2, 4/-2)

(x, y) = (1, -2)

Hence the point of intersection of two lines 2x+4y+6= 0 and 2x+3y+4= 0

is (1, -2).

Answer: Given Straight Line Equations are:

2x+4y+2=0 and 2x+3y+5=0 

Implies a₁ =2, b₁ = 4, c₁=2, a₂=2, b₂=3, c₂=5

Intersecting point can be calculated by using the point of intersection formula,

(x, y) = ((b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁), (c₁a₂ – c₂a₁) / (a₁b₂ – a₂b₁))

                         (x, y) = ((45 – 32) / (23 -24), (22 – 25) / (23 – 24))

                          (x, y) = ((20-6) / (6-8), (4 -10) / (6-8))

                          (x, y) = (14 / -2, -6 / -2)

                          (x, y) = (-7, 3)

Thus, the point of intersection of two given lines is (-7, 3).

Answer: A point to be a point of intersection it should satisfy both the lines 2x + 3y – 21 = 0, 

x + 2y – 13 = 0. 

Substituting (x, y) = (3, 5) in both the lines where x = 3 and y = 5

2(3) +3(5) -21 = 0

6 +15 -21 = 0

0=0 satisfied

Also,

3 + 2(5) – 13 = 0

3+10 -13 = 0

0 = 0 satisfied

Since the two equations are satisfied it is a point of intersection of both the lines.