Length of Intercept Cut-Off from a Line: Formula

The study of lines and their properties is a fundamental aspect of analytic geometry. One of the key concepts in this area is the "intercept," specifically the segments intercepted by a line on the x-axis and y-axis of a Cartesian coordinate system. These segments are known as the x-intercept and y-intercept, respectively. The lengths of these intercepts provide valuable geometric information about the line, such as its position relative to the axes and its slope.

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1. Length of Intercept Cut-Off from a line
2. Solved Example Based on Length of Intercept Cut-Off from a line:
3. Summary

Length of Intercept Cut-Off from a line

To understand the concept of intercepts, we begin with the standard form of the equation of a line: $a x+b y+c=0$

$\text { Here, $a$, b, and c are constants, and ( } x, y \text { ) represents the coordinates of any point on the line. }$. This equation can describe any straight line in a two-dimensional plane, provided that aaa and bbb are not both zero.

x-Intercept: The x-intercept is the point where the line crosses the x-axis. At this point, the y-coordinate is zero. To find the x-intercept, we set y=0 in the line equation and solve for x.

y-Intercept: The y-intercept is the point where the line crosses the y-axis. At this point, the x-coordinate is zero. To find the y-intercept, we set x=0 in the line equation and solve for y.

The length of the intercept cut off from the line $L: y=m x+c$ by the circle $x^2+y^2=a^2$ is

$2 \times \sqrt{\left(\frac{a^2\left(1+m^2\right)-c^2}{\left(1+m^2\right)}\right)}$

Proof:

$\mathrm{OM}=\left|\frac{\mathrm{c}}{\sqrt{1+\mathrm{m}^2}}\right|$
In $\Delta$ OAM,

$\begin{aligned}
\mathrm{AM}^2 & =\mathrm{AO}^2-\mathrm{OM}^2 \\
& =\mathrm{a}^2-\frac{\mathrm{c}^2}{1+\mathrm{m}^2} \\
& =\frac{\mathrm{a}^2\left(\mathrm{~m}^2+1\right)-\mathrm{c}^2}{\left(1+\mathrm{m}^2\right)}
\end{aligned}$
$\Rightarrow \quad \mathrm{AM}=\sqrt{\frac{\mathrm{a}^2\left(\mathrm{~m}^2+1\right)-\mathrm{c}^2}{\left(1+\mathrm{m}^2\right)}}$
The length of the intercept is $=A B$

$\begin{gathered}
=2 \mathrm{AM} \\
\mathrm{AB}=2 \times \sqrt{\frac{\mathrm{a}^2\left(\mathrm{~m}^2+1\right)-\mathrm{c}^2}{\left(1+\mathrm{m}^2\right)}}
\end{gathered}$

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Solved Example Based on Length of Intercept Cut-Off from a line:

Example 1: The sum of the squares of the lengths of the chords intercepted on the circle, $x^2+y^2=16$, by the lines $x+y=n, n \in N$, where $N$ is the set of all natural numbers, is:

1) 105

2) 210

3) 160

4) 320

Solution

The length of intercept cut-off from the line $y=m x+c_{\text {to the circle }} x^2+y^2=a^2$ is

$\sqrt{a^2+\frac{c^2}{\left(1+m^2\right)}}$
Perpendicular distance of a point from a line -

$\rho=\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}$

- wherein
$\rho$ is the distance from the line $a x+b y+c=0$.
Summation of series of natural numbers -

$\sum_{k=1}^n K^2=\frac{1}{6} n(n+1)(2 n+1)$

- wherein

The sum of squares of first $n$ natural numbers

$\begin{aligned}
& 1^2+2^2+3^2+4^2+------+n^2=\frac{n(n+1)(2 n+1)}{6} \\
& x^2+y^2=16 \\
& x+y=n
\end{aligned}$

then length of perpendicular from centre $(0,0)$ to line $x+y=n=\left|\frac{0+0-n}{\sqrt{1^2+1^2}}\right|=\frac{n}{\sqrt{2}}$

$\begin{aligned}
& \text { length of intercepts }=\sqrt[2]{4^2-\frac{n^2}{(\sqrt{2})^2}}=2 \sqrt{16-\frac{n^2}{2}} \\
& =\sqrt{64-2 n^2}
\end{aligned}$

Possible values of n are $=1,2,3,4,5$
Sum of squares of length $\sum_{i=1}^5\left(\sqrt{64-2 x^2}\right)^2$

$\begin{aligned}
& =\sum_{i=1}^5\left(64-2 x^2\right) \\
& =64 \times 5-2 \sum_{i=1}^5 n^2 \\
& =64 \times 5-2 \frac{n(n+1)(2 n+1)}{6} \\
& =320-2 \frac{5 \times 6 \times 11}{6}
\end{aligned}$
$\begin{aligned}
& =320-110 \\
& =210
\end{aligned}$
Example 2: If the length of the chord of the circle $x^2+y^2=r^2(r>0)$ along the line $y-2 x=3$ is r then $r^2$ is equal to
1) $\frac{9}{5}$
2) 12
3) $\frac{12}{5}$
4) $\frac{24}{5}$

Solution

Length of the chord by the line $y=m x+c$ on the circle $x^2+y^2=a^2$ is

$=2 \sqrt{\frac{a^2\left(1+m^2\right)-c^2}{1+m^2}}$

Given the equation of the line, $y=2 x+3$
and circle $x^2+y^2=r^2$
according to question

$r=2 \sqrt{\frac{r^2(1+4)-9}{1+4}} \Rightarrow \frac{r^2}{4}=\frac{5 r^2-9}{5} \Rightarrow r^2=\frac{12}{5}$

Hence, the answer is the option 3.

Example 3: Let a circle $C:(x-h)^2+(y-k)^2=r^2, k>0$, touch the $x$-axis at $(1,0)$. If the line $x+y=0$ intersects the circle $C$ at $P$ and $Q$ such that the length of the chord $P Q$ is 2 . then the value of $h+k+r$ is equal to
1) 7
2) 6
3) 9
4) 4

Solution

$\begin{aligned}
& (x-h)^2+(y-k)^2=r^2 ; k>0 \\
& (x-1)^2+y^2-2 k y+k^2=k^2 \\
& x^2+y^2-2 x-2 k y+1=0 \\
& k^2-\left(\frac{1+k}{\sqrt{2}}\right)^2=1 \\
& 2 k^2-1-k^2-2 k=2 \\
& k^2-2 k-3=0 \\
& k=3,-1 \\
& h+k+r=1+3+3=7
\end{aligned}$
Hence, the answer is 7

Example 4: The equation of the straight line passing through the point $(4,3)$ and making intercept on the co-ordinates axes whose sum is -1 , is
1) $\frac{x}{2}-\frac{y}{3}=-1$ and $\frac{x}{2}+\frac{y}{1}=1$
2) $\frac{x}{2}-\frac{y}{3}=-1$ and $\frac{x}{-2}+\frac{y}{1}=-1$
3) $\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{2}+\frac{y}{1}=1$
4) $\frac{x}{2}+\frac{y}{3}=-1 \quad$ and $\quad \frac{x}{-2}+\frac{y}{1}=-1$

Solution
Let the equation of line is $\frac{x}{a}+\frac{y}{-1-a}=1$, which passes through $(4,3)$.
Hence equation is $\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$
Hence, the answer is the option 4.

Summary

The length of the intercept cut-off from a line on the coordinate axes provides crucial information about the line's position and orientation relative to the axes. Understanding how to calculate these intercepts and their lengths is essential in geometry, mathematics, and various practical applications, including engineering, computer graphics, and optimization problems. By mastering these concepts, one can solve a range of geometric problems and gain insights into the spatial relationships between lines and coordinate systems.