Find the equation of the lines which passes through the point (3,4) and cuts off intercepts from the coordinat

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(3,4)

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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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Two lines satisfy the conditions: $x + y = 7$ and $4x + 3y = 24$ .

Let the line make intercepts $a$ and $b$ on the axes, so its equation is

\frac{x}{a} + \frac{y}{b} = 1,\qquad a + b = 14 \ \Rightarrow\ b = 14 - a.

Since it passes through $(3, 4)$ :

\frac{3}{a} + \frac{4}{14-a} = 1.

Clearing denominators:

3(14-a) + 4a = a(14-a)\ \Rightarrow\ 42 + a = 14a - a^2\ \Rightarrow\ a^2 - 13a + 42 = 0.

$(a-6)(a-7) = 0 \ \Rightarrow\ a = 6 \ \text{or}\ a = 7.$ …

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

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL2 marks
$Q. Obtain the equation of a straight line in the intercept form.$
›Reveal solutionSolution

A line meeting the axes at $(a,0)$ and $(0,b)$ has equation $\dfrac xa+\dfrac yb=1$ — the intercept form.

Let the line cut the $x$ -axis at $(a,0)$ and the $y$ -axis at $(0,b)$ .

Step 1: Use the two-point form of a line.

For a line through $(x_1,y_1)$ and $(x_2,y_2)$ :

$y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

With $(x_1,y_1)=(a,0)$ and $(x_2,y_2)=(0,b)$ :

$y-0 = \dfrac{b-0}{0-a}(x-a) = -\dfrac{b}{a}(x-a)$

Step 2: Simplify.

$ay = -b(x-a) = -bx+ab$ …

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL2 marks
$(a,b)$
›Reveal solutionSolution

The line meets the axes at $(2a,0)$ and $(0,2b)$ (since $P(a,b)$ is their midpoint), and its intercept-form equation simplifies to $x/a+y/b=2$ .

Let the line meet the $x$ -axis at $(X,0)$ and the $y$ -axis at $(0,Y)$ . The midpoint of these two intercept points is

$\left(\dfrac{X}{2},\dfrac{Y}{2}\right).$

Since $P(a,b)$ is this midpoint:

$\dfrac{X}{2}=a\ \Rightarrow\ X=2a,\qquad \dfrac{Y}{2}=b\ \Rightarrow\ Y=2b.$

…

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Q.Find the equation of the lines which passes through the point (3,4)(3,4)(3,4) and cuts off intercepts from the coordinate axes such that their sum is 14.

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Q.Find the equation of the lines which passes through the point $(3,4)$ and cuts off intercepts from the coordinate axes such that their sum is 14.

Unlock everything free for 14 days