Question

7. A variable line \( L_{1} \) cuts \( y=3 x+1 \) and \( y=-2 x+3 \) at points \( P_{1} \) and \( P_{2} \). If the locus of midpoints of \( P_{1} \) and line \( L_{2} \) with underfined slope where slope of \( L_{1} \) is constant. If slope of \( L_{1} \) is \( p / q \), where \( p, q \) are \( c 0 \) ? natural number, then find \( p+q \).

Answer 1

Let equation of L₁ is y = p/q x + c

For the x-coordinate of p₁ we must

Solve 

y = 3x + 1

△ y = p/q x + c

after solving we have \(x=\frac{1-c}{\frac{p}{q}-3}\)

Now for x-coordinate of p₂ we must

Solve

y = -2x + 3

y = p/q x + c

after solving we have \(x=\frac{3-c}{2+\frac{p}{q}}\)

Now, x-coordinate of M i.e mid-point of p₁ and p₂ can be given as

\(x=\frac{\frac{1-c}{\frac{p}{q}-3}+\frac{3-c}{2+\frac{p}{q}}}{2}\) ---(1)

But M lies on the line L₂ whose slope is undefined hence its equation is x = constant

If equation (1) is constant then

we must have dx/dc = 0 as p/q is also constant.

⇒ dx/dc = 0

⇒ \(\frac{1}{(\frac{p}{q}-3)}\times(-1)+\frac{1}{2+\frac{p}{q}}\times (-1)= 0\)

⇒ \(\frac{2+\frac{p}{q}+\frac{p}{q}-3}{(\frac{p}{q}-3)\,(2+\frac{p}{q})}\)

⇒ \(2\frac{p}{q}=1\)

⇒ \(\frac{p}{q}=\frac{1}{2}\)

⇒ p + q = 3