Question

A line is such that its segment
between the lines
5x-y+4= 0 and 3x + 4y - 4 = 0 is bisected at the
point (1,5). Obtain its equation.​

Answer 1

Solution:

The Equations of lines given are:

• 5x - y + 4 = 0
• 3x + 4y - 4 = 0

Let

• A (α₁, β₁) and B (α₂, β₂) be the line segment in between these 2 lines and
• P (1, 5) be the midpoint of 2 lines

Applying midpoint formula,

$\sf{midpoint \: of \: AB = \bigg( \dfrac{ \alpha_{1} + \alpha_{2} }{2} ,\dfrac{ \beta_{1} + \beta_{2}}{2} \bigg) }$

$\implies \sf{(1,5) = \bigg(\dfrac{\alpha_{1} + \alpha_{2} }{2} ,\dfrac{ \beta_{1} + \beta_{2}}{2} \bigg) }$

Equating X-coordinates,

$\dfrac{ \alpha_{1} + \alpha_{2} }{2} = 1$

$\implies \sf{ \alpha_{1} + \alpha_{2} = 2}$

$\implies \sf{ \alpha_{2} = 2 - \alpha_{1}}$

Equating Y-coordinates,

$\dfrac{ \beta_{1} + \beta_{2}}{2} = 5$

$\implies \sf{ \beta_{1} + \beta_{2} = 10}$

$\implies \sf{ \beta_{2} = 10 - \beta_{1}}$

Point A lies on line 1.

So, (α₁, β₁) satisfies the equation of line 1.

5x - y + 4 = 0

→ 5α₁ - β₁ + 4 = 0 ----- [Equation 1]

Similarly, Point B lies on line 2.

So, (α₂, β₂) satisfies the equation of line 2.

3x + 4y - 4 = 0

→ 3α₂ + 4β₂ - 4 = 0

Now substitute the value of α₂ and β₂ from above.

3α₂ + 4β₂ - 4 = 0

→ 3(2 - α₁) + 4(10 - β₁) - 4 = 0

→ 6 - 3α₁ + 40 - 4β₁ - 4 = 0

→ 42 - 3α₁ - 4β₁ = 0

→ 42 = 3α₁ + 4β₁

→ 3α₁ + 4β₁ - 42 = 0 ----- [Equation 2]

Now we have 2 equations in terms of α₁ and β₁.

Multiply Equation 1 by 4.

4 (5α₁ - β₁ + 4) = 4(0)

→ 20α₁ - 4β₁ + 16 = 0 ----- [Equation 3]

Adding Equation 2 and Equation 3,

3α₁ + 4β₁ - 42 = 0

{+} 20α₁ - 4β₁ + 16 = 0

23α₁ - 26 = 0

→ 23α₁ = 26

→ α₁ = 26/23

Substitute the value of α₁ in Equation 1 to find the value of β₁.

5α₁ - β₁ + 4 = 0

$\implies \sf{5 \bigg( \dfrac{26}{23} \bigg) - \beta_{1} + 4 = 0 }$

$\implies \sf{ \beta_{1} = 5 \bigg( \dfrac{26}{23} \bigg) + 4 }$

$\implies \sf{ \beta_{1} = \dfrac{130}{23} + 4 }$

$\implies \sf{\beta_{1} = \dfrac{130 + 92}{23} }$

$\implies \bf{\beta_{1} = \dfrac{222}{23} }$

Hence, AB passes through the points P (1, 5) and (26/23, 222/23)

We know the equation of line joining 2 points as:

$\boxed{ \boldsymbol{y -y_{1} = \dfrac{y_{2} -y_{1}}{x_{2} - x_{1}} (x - x_{1})}}$

Substituting the values,

$\implies \sf{y-5 =\dfrac{\dfrac{222}{23} - 5}{\dfrac{26}{23}-1 } \: \: \big(x - 1 \big) }$

On solving this,

$\implies \sf{y-5=\frac{107}{3}(x-1)}$

⇒ 3y - 15 = 107x - 107

⇒ 3y - 107x - 15 + 107 = 0

⇒ -107x + 3y + 92 = 0

⇒ 107x - 3y - 92 = 0

∴ Equation of line = 107x - 3y - 92 = 0