Find point (or points) which are at distance of $\sqrt{10}$ units from the point (4, 3) given that the ordinate of the point (or points) is twice the abscissa.

Given,

Ordinate of the point is twice the abscissa.

Let abscissa of point be k, then ordinate = 2k

Let the required point be P.

∴ P = (k, 2k).

By distance formula,

d = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Given,

Distance between (4, 3) and (k, 2k) = $\sqrt{10}$ units.

$\Rightarrow \sqrt{10} = \sqrt{(k - 4)^2 + (2k - 3)^2} \\[1em] \Rightarrow \sqrt{10} = \sqrt{k^2 + (4)^2 - 8k + (2k)^2 + (3)^2 - 12k} \\[1em] \Rightarrow \sqrt{10} = \sqrt{k^2 + 16 - 8k + 4k^2 + 9 - 12k}$

On squaring both sides

$\Rightarrow 10 = k^2 + 16 - 8k + 4k^2 + 9 - 12k \Rightarrow 10 = 5k^2 - 20k + 25 \\[1em] \Rightarrow 10 = 5(k^2 - 4k + 5) \\[1em] \Rightarrow k^2 - 4k + 5 = 2 \\[1em] \Rightarrow k^2 - 4k + 5 - 2 = 0 \\[1em] \Rightarrow k^2 - 4k + 3 = 0 \\[1em] \Rightarrow k^2 - 3k - k + 3 = 0 \\[1em] \Rightarrow k(k - 3) - 1(k - 3) = 0 \\[1em] \Rightarrow (k - 1)(k - 3) = 0 \\[1em] \Rightarrow k - 1 = 0 \text{ or } k - 3 = 0 \\[1em] \Rightarrow k = 1 \text{ or } k = 3.$

When k = 1,

P = (k, 2k) = (1, 2).

When k = 3,

P = (k, 2k) = (3, 6).

Hence, required points are (1, 2) or (3, 6).