Question

the locus of hte point,for which the sum of the distances from the coordinate axes is 9 is

Answer 1

Required locus: $|x|+|y|=9$

Concept to be used:

The distance of the point $(x_{1},y_{1})$ from the straight line $ax+by+c=0$ is

$d=\dfrac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}}$

Step-by-step explanation:

Step 1 of 5.

Let, the required point be $(h,k)$.

Step 2 of 5.

The equation of the x-axis is

$y=0$

$\Rightarrow 0.x+1.y=0$

Then the distance of the point $(h,k)$ from the x-axis is

$d_{1}=\dfrac{|k|}{\sqrt{0^{2}+1^{2}}}$

$\Rightarrow d_{1}=|k|$

Step 3 of 5.

The equation of the y-axis is

$x=0$

$\Rightarrow 1.x+0.y=0$

Then the distance of the point $(h,k)$ from the y-axis is

$d_{2}=\dfrac{|h|}{\sqrt{1^{2}+0^{2}}}$

$\Rightarrow d_{2}=|h|$

Step 4 of 5.

Given that, the sum of the distances $d_{1}$ and $d_{2}$ is 9.

Then, $d_{1}+d_{2}=9$

$\Rightarrow |k|+|h|=9$

$\Rightarrow |h|+|k|=9$ ... (1)

Step 5 of 5.

In order to find the required locus, we write $h=x$ and $k=y$ in (1) no. equation and we get the required locus,

$|x|+|y|=9$

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