The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and

(a) Given equations are 2x – 3y = 0 …(i)

and 4x – 5y = 2 …(ii)

Equation of line passing through eq. (i) and (ii), we get

(2x – 3y) + λ(4x – 5y – 2 ) = 0 …(iii)

If the above eq. passes through the point (2, 1), we get

(2 × 2 – 3 × 1) + λ (4 × 2 – 5 × 1 – 2) = 0

⇒ (4 – 3) + λ (8 – 5 – 2) = 0

⇒ 1 + λ = 0

⇒ λ = -1

Putting the value of λ in eq. (iii), we get

(2x – 3y) + (-1)(4x – 5y – 2 ) = 0

⇒ 2x – 3y – 4x + 5y + 2 = 0

⇒ -2x + 2y + 2 = 0

⇒ x – y – 1 = 0

Hence, (a) ↔ (iii)

(b) Given equations are 2x – 3y = 0 …(i)

and 4x – 5y = 2 …(ii)

Equation of line passing through eq. (i) and (ii), we get

(2x – 3y) + λ(4x – 5y – 2 ) = 0 …(iii)

⇒ 2x – 3y + 4λx - 5λy - 2λ = 0

⇒ x(2 + 4λ) – y(3 + 5λ) – 2λ = 0

⇒ – y(3 + 5λ) = -(2 + 4λ) + 2λ

Since, the above equation is in y = mx + b form.

So, Slope of eq. (iii) is

Now, we find the slope of the given line x + 2y + 1 = 0 …(iv)

⇒ 2y = -x – 1

Since, the above equation is in y = mx + b form.

So, Slope of eq. (iv) is

We know that, if two lines are perpendicular to each other than the product of their slopes is equal to -1 i.e.

m₁m₂ = -1

Now, given that eq. (iii) is perpendicular to the given line x + 2y + 1 = 0

⇒ 2 + 4λ = 2 × (3 + 5λ)

⇒ 2 + 4λ = 6 + 10λ

⇒ 4λ – 10λ = 6 – 2

⇒ -6λ = 4

Putting the value of λ in eq. (iii), we get

⇒ 6x – 9y – 8x + 10y + 4 = 0

⇒ -2x + y + 4 =0

⇒ 2x – y – 4 = 0

⇒ 2x – y = 4

Hence, (b) ↔ (i)

(c) Given equations are 2x – 3y = 0 …(i)

and 4x – 5y = 2 …(ii)

Equation of line passing through eq. (i) and (ii), we get

(2x – 3y) + λ(4x – 5y – 2 ) = 0 …(iii)

⇒ 2x – 3y + 4λx - 5λy - 2λ = 0

⇒ x(2 + 4λ) – y(3 + 5λ) – 2λ = 0

⇒ – y(3 + 5λ) = -(2 + 4λ) + 2λ

Since, the above equation is in y = mx + b form.

So, Slope of eq. (iii) is

Now, we find the slope of the given line 3x – 4y + 5 = 0

⇒ 3x + 5 = 4y

Since, the above equation is in y = mx + b form.

So, the Slope of above equation is

Now, we know that if the two lines are parallel than there slopes are equal.

∴ m₁ = m₂

⇒ 4(2 + 4λ) = 3(3 + 5λ)

⇒ 8 + 16λ = 9 + 15λ

⇒ 16λ – 15λ = 9 – 8

⇒ λ = 1

Putting the value of λ in eq. (iii), we get

(2x – 3y) + (1)(4x – 5y – 2 ) = 0

⇒ 2x – 3y + 4x – 5y – 2 = 0

⇒ 6x – 8y – 2 = 0

⇒ 3x – 4y – 1 = 0

Hence (c) ↔ (iv)

(d) Given equations are 2x – 3y = 0 …(i)

and 4x – 5y = 2 …(ii)

Equation of line passing through eq. (i) and (ii), we get

(2x – 3y) + λ(4x – 5y – 2 ) = 0 …(iii)

⇒ 2x – 3y + 4λx - 5λy - 2λ = 0

⇒ x(2 + 4λ) – y(3 + 5λ) – 2λ = 0

⇒ – y(3 + 5λ) = -(2 + 4λ) + 2λ

Since, the above equation is in y = mx + b form.

So, Slope of eq. (iii) is

Given that the equation is equally inclined with axes, it means that the lines make the equal angles with both the coordinate axes. It will make an angle of 45° or 135° with x – axis.

So, the slope of two lines equally inclined to the axes are

m₂ = tan 45° and tan 135°

= 1 and tan (180 – 45°)

= 1 and -1

So,

⇒ 2 + 4λ = - (3 + 5λ) ⇒ 2 + 4λ = (3 + 5λ)

⇒ 2 + 4λ = - 3 – 5λ ⇒ 4λ – 5λ = 3 – 2

⇒ 4λ + 5λ = -3 – 2 ⇒ -λ = 1

⇒ 9λ = -5 ⇒ λ = -1

Putting the value of λ in eq. (iii), we get

⇒ 18x – 27y – 20x + 25y + 10 = 0

⇒ -2x – 2y + 10 = 0

⇒ x + y – 5 = 0

If λ = -1, then the required equation is

(2x – 3y) + (-1)(4x – 5y – 2) = 0

⇒ 2x – 3y – 4x + 5y + 2 = 0

⇒ -2x + 2y + 2 = 0

⇒ x – y – 1 = 0