Reduce the equation 3x – 4y + 12 = 0 to intercepts form.

Question:

Reduce the equation 3x – 4y + 12 = 0 to intercepts form. Hence, find the length of the portion of the line intercepted between the axes

 

Solution:

Given equation is $3 x-4 y+12=0$

We can rewrite it as $3 x-4 y=-12$

$\Rightarrow \frac{3}{-12} \mathrm{x}+\frac{4}{12} \mathrm{y}=1$

$\Rightarrow \frac{\mathrm{x}}{-4}+\frac{\mathrm{y}}{3}=1$

This equation is in the slope intercept form i.e. in the form

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

Where, $x$-intercept $=-4$ and $y$-intercept $=3$

Two points are: $(-4,0)$ on the $x$-axis and $(0,3)$ on $y$-axis

We know distance between two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ is

$=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

Length of the line

$=\sqrt{(-4-0)^{2}+(0-3)^{2}}$

$=\sqrt{16+9}$

$=\sqrt{25}$

$=5$ 

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