If the line $\dfrac{x}{a}+\dfrac{y}{b}=1$ passes through the points $(2,-3)$ and $(4,-5)$, then $(a,b)$ is (A) $(1,1)$ (B) $(-1,1)$ (C) $(1,-1)$ (D) $(-1,-1)$

If the line (x)/(a)+(y)/(b)=1 passes through the points (2,-3) and (4,-5), then (a,b) is (A) (1,1) (B) (-1,1)

NCERT Exemplar · Q27

\dfrac{x}{a}+\dfrac{y}{b}=1

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Concept understanding — Coordinate Axes Equations

Coordinate Axes Equations

Imagine a blank sheet of paper. You want to describe exactly where a point is. The most natural way is to draw two perpendicular lines — one horizontal, one vertical — and measure distances from them. Those two lines are the coordinate axes.

The horizontal line is the x-axis. The vertical line is the y-axis. Where they cross is the origin, the point $(0,0)$ .

Now, here's the key idea: every point on the x-axis has a y-coordinate of 0. Why? Because to be on that horizontal line, you haven't moved up or down at all — your vertical distance from the axis is zero. Similarly, every point on the y-axis has an x-coordinate of 0 — you haven't moved left or right from the vertical line.

That's the entire intuition. Let's make it precise.

The Equations

\text{x-axis: } y = 0

That's it. These are the equations of the coordinate axes.

$y = 0$ means: "all points where the y-coordinate is zero, regardless of x." That's the entire horizontal line through the origin.
$x = 0$ means: "all points where the x-coordinate is zero, regardless of y." That's the entire vertical line through the origin.

Watch out

A common mistake is to think $y=0$ means "the point $(0,0)$ ". No — it means every point with $y=0$ , like $(5,0)$ , $(-3,0)$ , $(100,0)$ , etc. It's a whole line, not a single point.

Why This Matters

These two equations are the foundation for everything in coordinate geometry. Every other line, curve, or shape is described relative to these axes. For example:

The line $y = 2$ is a horizontal line parallel to the x-axis, shifted up by 2 units.
The line $x = -3$ is a vertical line parallel to the y-axis, shifted left by 3 units.

So when you see $y = 0$ or $x = 0$ , think: "the original, unshifted axes themselves." …

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If a line passes through given points, their coordinates must satisfy the line's equation. Substituting the two points into the equation $\frac{x}{a} + \frac{y}{b} = 1$ yields a system of two linear equations in terms of $\frac{1}{a}$ and $\frac{1}{b}$ . Solving this system gives $\frac{1}{a} = -1$ and $\frac{1}{b} = -1$ , so $(a,b) = (-1,-1)$ .

The problem asks us to find the values of $a$ and $b$ for a line given in intercept form, $\frac{x}{a} + \frac{y}{b} = 1$ , which passes through two specific points.

The fundamental concept here is that if a point lies on a line, its coordinates must satisfy the equation of that line. This means that if we substitute the $x$ and $y$ coordinates of a point into the line's equation, the equation must hold true. Since the line passes through two points, we can perform this substitution twice, generating two separate equations. These two equations will form a system that we can then solve to find the unknown values of $a$ and $b$ .

Let's apply this concept step-by-step.

Formulate equations from the given points.

The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$ .

The line passes through $(2, -3)$ and $(4, -5)$ .
- Using the point $(2, -3)$ :
  
  Substitute $x=2$ and $y=-3$ into the line equation:
  
  $\frac{2}{a} + \frac{-3}{b} = 1$
  
  This simplifies to:
  
  $\frac{2}{a} - \frac{3}{b} = 1 \quad \text{(Equation 1)}$
- Using the point $(4, -5)$ :
  
  Substitute $x=4$ and $y=-5$ into the line equation:
  
  $\frac{4}{a} + \frac{-5}{b} = 1$
  
  This simplifies to:
  
  $\frac{4}{a} - \frac{5}{b} = 1 \quad \text{(Equation 2)}$
Solve the system of linear equations.

We now have a system of two equations with two unknowns, $\frac{1}{a}$ and $\frac{1}{b}$ :
1. $2\left(\frac{1}{a}\right) - 3\left(\frac{1}{b}\right) = 1$
2. $4\left(\frac{1}{a}\right) - 5\left(\frac{1}{b}\right) = 1$
To make this system look more familiar, let's substitute $X = \frac{1}{a}$ and $Y = \frac{1}{b}$ .
1. $2X - 3Y = 1$
2. $4X - 5Y = 1$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$(x, y, z)$
›Reveal solutionSolution
The sum of the squares of perpendicular distances from a point to the coordinate axes equals twice the square of the distance from the origin, so $k = 2$ .

We are asked: For a point $(x, y, z)$ , the sum of the squares of its perpendicular distances to the three coordinate axes equals $k$ times the square of its distance from the origin. Find $k$ .

Concept & Intuition

The "perpendicular distance" from a point to an axis is the shortest distance from the point to any point on that axis. For the $x$ -axis, the foot of the perpendicular is $(x, 0, 0)$ , so the distance is $\sqrt{y^2 + z^2}$ . Similarly for the $y$ -axis and $z$ -axis. The distance from the origin is $\sqrt{x^2 + y^2 + z^2}$ . The problem asks how these quantities relate — specifically, the sum of the squares of those three axis-distances compared to the square of the origin-distance.

Step-by-step reasoning

Write the perpendicular distances to each axis

Distance to $x$ -axis: $\sqrt{y^2 + z^2}$

Distance to $y$ -axis: $\sqrt{x^2 + z^2}$

Distance to $z$ -axis: $\sqrt{x^2 + y^2}$

Square each distance

Square of distance to $x$ -axis: $y^2 + z^2$

Square of distance to $y$ -axis: $x^2 + z^2$

Square of distance to $z$ -axis: $x^2 + y^2$

Sum these squares

$(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 2x^2 + 2y^2 + 2z^2$

Compare with the square of the distance from the origin …
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Q.If the line xa+yb=1\dfrac{x}{a}+\dfrac{y}{b}=1ax​+by​=1 passes through the points (2,−3)(2,-3)(2,−3) and (4,−5)(4,-5)(4,−5), then (a,b)(a,b)(a,b) is (A) (1,1)(1,1)(1,1) (B) (−1,1)(-1,1)(−1,1) (C) (1,−1)(1,-1)(1,−1) (D) (−1,−1)(-1,-1)(−1,−1)

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Q.If the line $\dfrac{x}{a}+\dfrac{y}{b}=1$ passes through the points $(2,-3)$ and $(4,-5)$ , then $(a,b)$ is
(A) $(1,1)$
(B) $(-1,1)$
(C) $(1,-1)$
(D) $(-1,-1)$

Unlock everything free for 14 days