The equation of the straight line passing through the point (4, 3) and making intercepts on the co-o

Q: Why can't we just use the slope-intercept form $y = mx + c$?

You absolutely could, but it would be like using a sledgehammer to crack a nut. If you use $y = mx + c$, you have two unknowns, $m$ and $c$. You would need to express $c$ in terms of $m$ using the point $(4, 3)$, and then relate the intercepts to $m$ and $c$. It is algebraically messy and prone to errors. The intercept form $\frac{x}{a} + \frac{y}{b} = 1$ is elegant because it maps directly to the language of the problem. Always choose the tool that fits the geometry!

Q: Why are there two answers? Did I do something wrong?

On the contrary, finding two answers is a sign that you have mastered the problem! In coordinate geometry, when you encounter a quadratic equation, it often signifies that there are multiple geometric configurations that satisfy the given conditions. Here, the quadratic nature of the equation arises because the relationship between the intercepts and the point is non-linear. Both lines are perfectly valid solutions.

Q: What if the line is vertical or horizontal? Does the intercept form still work?

That is a brilliant question. The standard intercept form $\frac{x}{a} + \frac{y}{b} = 1$ assumes that the line is not parallel to either axis (i.e., $a$ and $b$ are finite and non-zero). If the line were vertical or horizontal, the intercepts would be undefined or zero. However, in this problem, the point $(4, 3)$ is in the first quadrant, and the sum of intercepts is $-1$. A vertical or horizontal line passing through $(4, 3)$ would have intercepts that do not sum to $-1$. Thus, we are safe to use the intercept form.

Question

The equation of the straight line passing through the point $(4, 3)$ and making intercepts on the co-ordinate axes whose sum is $-1$ is

Options:

(A) $\frac{x}{2} - \frac{y}{3} = 1$ and $\frac{x}{-2} + \frac{y}{1} = 1$

(B) $\frac{x}{2} - \frac{y}{3} = -1$ and $\frac{x}{-2} + \frac{y}{1} = -1$

(C) $\frac{x}{2} + \frac{y}{3} = 1$ and $\frac{x}{-2} + \frac{y}{-1} = 1$

(D) $\frac{x}{2} + \frac{y}{-3} = -1$ and $\frac{x}{-2} + \frac{y}{1} = -1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to the beautiful world of coordinate geometry. Today, we are not just solving an equation; we are exploring the dance of a line as it pivots through space.

Imagine you are standing on a vast, infinite grid holding a straight rod. You are told that this rod must pass through a specific, fixed point: $P(4, 3)$.

However, the rod must obey a strict rule: the sum of the distances where it cuts the $x$-axis and the $y$-axis must be exactly $-1$. This is a puzzle of constraints that we shall now solve.

### The Power of the Intercept Form

When you see the word 'intercepts' in a JEE problem, your mind should immediately light up. Do not reach for $y = mx + c$, as that is a tool for slopes.

Instead, reach for the intercept form:

$$\frac{x}{a} + \frac{y}{b} = 1$$

Here, $a$ is the $x$-intercept, and $b$ is the $y$-intercept. By setting this as our foundation, we are already halfway to the solution.

### The Constraint: The Leash on Our Line

The problem gives us a beautiful, simple constraint: the sum of the intercepts is $-1$. Mathematically, this is expressed as:

$$a + b = -1$$

This is the 'leash' that prevents the line from being just any line. We can rewrite this as $b = -1 - a$.

We have taken a problem with two unknowns, $a$ and $b$, and reduced it to a single variable, $a$. This is the essence of mathematical elegance: reducing complexity until the answer reveals itself.

### The Intersection of Point and Line

Now, we bring in our fixed point, $P(4, 3)$. Since our line must pass through this point, the coordinates $(4, 3)$ must satisfy our equation.

Substituting these into our intercept form, we get:

$$\frac{4}{a} + \frac{3}{b} = 1$$

We replace $b$ with $-1 - a$ to obtain:

$$\frac{4}{a} + \frac{3}{-1 - a} = 1$$

### The Algebraic Battle

To solve this, we find a common denominator for the left side:

$$\frac{4(-1 - a) + 3a}{a(-1 - a)} = 1$$

Expanding the numerator gives $-4 - 4a + 3a$, which simplifies to $-4 - a$. The denominator is $-a - a^{2}$. Thus, we have:

$$\frac{-4 - a}{-a - a^{2}} = 1$$

Cross-multiplying yields $-4 - a = -a - a^{2}$. The $-a$ terms on both sides cancel out perfectly, leaving us with:

$$-4 = -a^{2} \quad \Rightarrow \quad a^{2} = 4$$

This is the 'Aha!' moment. The quadratic nature of the equation reveals two possible values for $a$: $a = 2$ and $a = -2$.

### The Dual Reality

We have found two distinct values for $a$, meaning there are two distinct lines that satisfy our conditions.

**Case 1:** If $a = 2$, then $b = -1 - 2 = -3$. The equation of our first line is:

$$\frac{x}{2} - \frac{y}{3} = 1$$

**Case 2:** If $a = -2$, then $b = -1 - (-2) = 1$. The equation of our second line is:

$$\frac{x}{-2} + \frac{y}{1} = 1$$

We have navigated the constraints, performed the algebra, and uncovered the two geometric realities that satisfy the problem. Remember, in JEE Advanced, the math is about understanding the structure of the problem; you have seen how a quadratic equation can represent multiple geometric solutions.

Animated Solution for Mathematics - Straight Lines: The equation of the straight line passing through the point $(4, 3)$ and making intercepts on the co-ordinate axes whose sum is $- 1$ is

Select Answer:

Visualized Solution

Visualizing the Given Point

The Intercept Form

Applying the Sum Condition

Substituting the Point $(4, 3)$

Creating a Single Variable Equation

Simplifying the Equation

Solving the Quadratic Equation

Case 1: When $a = 2$

Case 2: When $a = - 2$

Final Conclusion

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Power of the Intercept Form

The Constraint

The Intersection of Point and Line

The Algebraic Battle

The Dual Reality

Similar Questions

A straight line through the point $A (3, 4)$ is such that its intercept between the axes is bisected at $A$ . Its equation is

If a straight line passing through the point $P (- 3, 4)$ is such that its intercepted portion between the coordinate axes is bisected at $P$ , then its equation is :

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

Slope of a line passing through $P (2, 3)$ and intersecting the line, $x + y = 7$ at a distance of 4 units from $P$ , is

Let $a, b, c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4 a x + 2 a y + c = 0$ and $5 b x + 2 b y + d = 0$ lies in the fourth quadrant and is equidistant from the two axes then

One side of a rectangle lies along the line $4 x + 7 y + 5 = 0$ . Two of its vertices are $(- 3, 1)$ and $(1, 1)$ . Find the equations of the other three sides.

A straight line $L$ is perpendicular to the line $5 x - y = 1$ . The area of the triangle formed by the line $L$ and the coordinate axes is 5. Find the equation of the line $L$ .

Let $P S$ be the median of the triangle with vertices $P (2, 2), Q (6, - 1)$ and $R (7, 3)$ . The equation of the line passing through $(1, - 1)$ and parallel to $P S$ is

The perpendicular bisector of the line segment joining $P (1, 4)$ and $Q (k, 3)$ has y-intercept $- 4$ . Then a possible value of $k$ is