Given: 3x – 4y + 5 = 0,

7x – 8y + 5 = 0

and 4x + 5y = 45

or 4x + 5y – 45 = 0

To show: Given lines are concurrent

The lines a₁x + b₁y + c₁ = 0, a₁x + b₁y + c₁ = 0 and a₁x + b₁y + c₁ = 0 are concurrent if

We know that,

We have,

a₁ = 3, b₁ = -4, c₁ = 5

a₂ = 7, b₂ = -8, c₂ = 5

a₃ = 4, b₂ = 5, c₃ = -45

Now, expanding along first row, we get

⇒ 3[(-8)(-45) – (5)(5)] – (-4)[(7)(-45) – (4)(5)] + 5[(7)(5) – (4)(-8)]

⇒ 3[360 – 25] + 4[-315 – 20] + 5[35 + 32]

⇒ 3[335] + 4[-335] + 5[67]

⇒ 1005 – 1340 + 335

⇒ 1340 – 1340

= 0

So, the given lines are concurrent.

Now, we have to find the point of intersection of the given lines

3x – 4y + 5 = 0,

7x – 8y + 5 = 0

and 4x + 5y – 45 = 0 …(A)

We know that, if three lines are concurrent the point of intersection of two lines lies on the third line.

So, firstly, we find the point of intersection of two lines

3x – 4y + 5 = 0, …(i)

7x – 8y + 5 = 0 …(ii)

Multiply the eq. (i) by 2, we get

6x – 8y + 10 = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

7x – 8y + 5 – 6x + 8y – 10 = 0

⇒ x – 5 = 0

⇒ x = 5

Putting the value of x in eq. (i), we get

3(5) – 4y + 5 = 0

⇒ 15 – 4y + 5 = 0

⇒ 20 – 4y = 0

⇒ -4y = -20

⇒ y = 5

Thus, the first two lines intersect at the point (5, 5). Putting x = 5 and y = 5 in eq. (A), we get

4(5) + 5(5) – 45

= 20 + 25 – 45

= 45 – 45

= 0

So, point (5, 5) lies on line 4x + 5y – 45 = 0

Hence, the point of intersection is (5, 5)