Length of Intercept Cut-Off from a Line: Formula

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.