**Question:**

Find the points on the X-axis which are at a distance of 2√5 from the point (7, -4). How many such points are there?

**Solution:**

We know that, every point on the X-axis in the form (x, 0). Let P(x, 0) the point on the X-axis

have 2√5 distance from the point Q (7, – 4).

By given condition, $\quad P Q=2 \sqrt{5} \quad\left[\because\right.$ distance formula $\left.=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$

$\Rightarrow \quad(P Q)^{2}=4 \times 5$

$\Rightarrow \quad(x-7)^{2}+(0+4)^{2}=20$

$\Rightarrow \quad x^{2}+49-14 x+16=20$

$\Rightarrow \quad x^{2}-14 x+65-20=0$

$\Rightarrow \quad x^{2}-14 x+45=0$

$\Rightarrow \quad x^{2}-9 x-5 x+45=0$ [by factorisation method]

$\Rightarrow \quad x(x-9)-5(x-9)=0$

$\Rightarrow \quad(x-9)(x-5)=0$

$x=5.9$

Hence, there are two points lies on the axis, which are (5, 0) and (9, 0), have 2V5 distance from the point (7, – 4).