Hey,Can anybody help me to deal with this question

Finding $h(x)$

A straight line through the origin making angle $\theta$ with the x-axis has equation

$y = (\tan\theta),x$

because slope is defined as $m = \tan\theta$ (rise over run in a right triangle).

Here $\theta = 60^\circ$, so

$h(x) = x\tan 60^\circ = \sqrt{3},x$

Finding $g(x)$

Two points are given: $(-a, 0)$ and $(0, a)$. Slope from two points:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{a - 0}{0 - (-a)} = \frac{a}{a} = 1$

Since $(0,a)$ is the y-intercept, using $y = mx + c$:

$g(x) = x + a$

Relating them

From $h(x) = \sqrt{3}x$ we get

$x = \frac{h(x)}{\sqrt{3}}$

Substitute into $g(x) = x + a$:

$g(x) = \frac{h(x)}{\sqrt{3}} + a$

Multiply both sides by $\sqrt{3}$:

$\sqrt{3},g(x) = h(x) + a\sqrt{3}$

$\boxed{\sqrt{3},g(x) - h(x) = a\sqrt{3}}$

Answer: (a)