What is the locus of the midpoint of PQ for lines through the intersection of 4x+3y−1=0 and 3x+4y−1=0?

x+y−14xy=0. Derived by writing the family of lines as L1+λL2=0, finding intercepts, setting midpoint=(h,k), and eliminating λ.

What is the point of intersection of 4x+3y=1 and 3x+4y=1?

(1/7, 1/7). Subtract the equations to get x=y, then solve.

Line through intersection of 4x+3y−1=0 and 3x+4y−1=0 meets axes at P and Q — locus of midpoint of PQ

JEERANKERS

MathCoord Geometry

QMCQCoordinate Geometry

A straight line drawn through the point of intersection of $4x+3y-1=0$ and $3x+4y-1=0$ meets the coordinate axes at points $P$ and $Q$. The locus of the midpoint of $PQ$ is:

A) $x+y-7=0$ B) $x+y-14xy=0$
C) $2x+y+14xy=0$ D) $x+2y-14xy=0$

✅ Correct Answer

B) x+y−14xy=0

✓

Solution

Write family of lines through the intersection

Any line through the intersection of $L_1:4x+3y-1=0$ and $L_2:3x+4y-1=0$ is:

$(4x+3y-1)+\lambda(3x+4y-1)=0$

$\Rightarrow\quad(4+3\lambda)x+(3+4\lambda)y=1+\lambda$

Find intercepts P and Q

x-intercept (set $y=0$): $P=\left(\dfrac{1+\lambda}{4+3\lambda},\ 0\right)$

y-intercept (set $x=0$): $Q=\left(0,\ \dfrac{1+\lambda}{3+4\lambda}\right)$

Let midpoint be (h, k) and eliminate λ

$h=\dfrac{1+\lambda}{2(4+3\lambda)}\quad\Rightarrow\quad\dfrac{1}{2h}=\dfrac{4+3\lambda}{1+\lambda}=3+\dfrac{1}{1+\lambda}$

$k=\dfrac{1+\lambda}{2(3+4\lambda)}\quad\Rightarrow\quad\dfrac{1}{2k}=\dfrac{3+4\lambda}{1+\lambda}=4-\dfrac{1}{1+\lambda}$

Adding the two results:

$$\frac{1}{2h}+\frac{1}{2k}=3+4=7\quad\Rightarrow\quad\frac{h+k}{2hk}=7\quad\Rightarrow\quad h+k=14hk$$

Replacing $(h,k)$ with $(x,y)$:

$$\boxed{x+y-14xy=0}$$

📘 Family of Lines & Intercept Locus

Any line through the intersection of $L_1=0$ and $L_2=0$ is written as $L_1+\lambda L_2=0$. For intercept problems, find $x$-intercept $(a,0)$ and $y$-intercept $(0,b)$, set midpoint $(h,k)=(a/2,b/2)$, express $1/h$ and $1/k$ in terms of the parameter, and eliminate the parameter to get the locus.

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Important Concepts

Family of Lines ConceptAny line through the intersection of $L_1=0$ and $L_2=0$ is $L_1+\lambda L_2=0$. This single equation with parameter $\lambda$ represents infinitely many lines through the fixed intersection point.

Intersection PointSolving $4x+3y=1$ and $3x+4y=1$: subtracting gives $x=y$, so $7x=1$, $x=y=1/7$. The family passes through $(1/7,\ 1/7)$.

Eliminating the ParameterKey step: $1/(2h)=3+1/(1+\lambda)$ and $1/(2k)=4-1/(1+\lambda)$. Adding eliminates the $1/(1+\lambda)$ term directly, giving $1/(2h)+1/(2k)=7$, i.e., $h+k=14hk$.

Geometric Meaning of Locus$x+y=14xy$ is a rectangular hyperbola (rewrite as $1/x+1/y=14$). It passes through all midpoints of segments cut by the family of lines on the coordinate axes.

FAQs

What is the point of intersection of the two given lines?

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Solving $4x+3y=1$ and $3x+4y=1$: subtract to get $x-y=0$, so $x=y$. Then $7x=1$, giving $x=y=1/7$. The intersection point is $(1/7,1/7)$.

How does adding 1/2h and 1/2k eliminate λ?

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$\frac{1}{2h}=3+\frac{1}{1+\lambda}$ and $\frac{1}{2k}=4-\frac{1}{1+\lambda}$. The $\frac{1}{1+\lambda}$ terms cancel on addition, directly giving $\frac{1}{2h}+\frac{1}{2k}=7$.

What is the curve x+y=14xy?

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It can be written as $\frac{1}{x}+\frac{1}{y}=14$, which is a rectangular hyperbola. In intercept form it shows that $1/x+1/y$ is always $14$ for any point on the locus.

How to verify: does the point (1,1) lie on the locus?

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$1+1-14(1)(1)=2-14=-12\neq0$. So $(1,1)$ does not lie on the locus, which makes sense.

Is this from JEE Main 2026?

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Yes, this question appeared in JEE Main 2026.

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