First determine the slope of the line $h$ perpendicular to $g(x):y=3x+2$ , where being perpendicular implies

$${\displaystyle m_2=-\frac{1}{m_1}}$$

Plug in the known values.

$${\displaystyle m_2=-\frac{1}{3}}$$

So the line we are looking for has a slope of $${\displaystyle -\frac{1}{3}}$$.

Now determine the $y$-axis intercept of the line $h$ through $P(3|5)$ by substituting the point $P$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+b$

Plug in the values.

$5=-\frac{1}{3}\cdot 3+b|+1$

Simplify and add $1$.

$6=b\iff b=6$

So the line equation is $${\displaystyle h(x):y=-\frac{1}{3}x+6}$$.

First determine the slope of the line $h$ perpendicular to $g(x):y=0.5x+1$ , where being perpendicular implies

$$m_2={\displaystyle -\frac{1}{m_1}}$$

Plug in the known values.

$${\displaystyle m_2=-\frac{1}{0.5}=-\frac{1}{\frac{1}{2}}=-2}$$

So the line we are looking for has a slope of $-2$.

Now determine the $y$-axis intercept of the line $h$ through $P(1|2)$ by substituting the point $P$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+b$

Plug in the values.

$2=-2\cdot 1+b$ $|+2$

Simplify and add 2.

$4=b\Rightarrow b=4$

So the line equation is $h(x):y=-2x+4$.

First determine the slope of the line $h$ perpendicular to $g(x):y=-5x+6$ , where being perpendicular implies

$${\displaystyle m_2=-\frac{1}{m_1}}$$

Plug in the known values.

$${\displaystyle m_2=-\frac{1}{-5}=0.2}$$

So the line we are looking for has a slope of $0.2$.

Now determine the $y$-axis intercept of the line $h$ through $P(-10|1)$ by substituting the point $P$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+b$

Plug in the values.

$1=0.2\cdot (-10)+b$ $|+2$

Simplify and add 2.

$3=b\Rightarrow b=3$

So the line equation is $h(x):y=0.2x+3$.

First determine the slope of the line $h$ perpendicular to $g(x):y=4x+3$ , where being perpendicular implies

$${\displaystyle m_2=-\frac{1}{m_1}}$$

Plug in the known values.

$$m_2=-{\displaystyle \frac{1}{4}=-0.25}$$

So the line we are looking for has a slope of $-0.25$.

Now determine the $y$-axis intercept of the line $h$ through $P(2|-5)$ by substituting the point $P$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+b$

Plug in the values.

$-5=-0.25\cdot 2+b$ $+0.5$

Simplify and add $0.5$.

$-4.5=b\Rightarrow b=-4.5$

So the line equation is $h(x):y=-0.25x-4.5$.

First determine the slope of the line $h$ perpendicular to $g(x):y=-\frac{2}{3}x+3$ , where being perpendicular implies

$${\displaystyle m_2=-\frac{1}{m_1}}$$

Plug in the known values.

$${\displaystyle m_2=-\frac{1}{-\frac{2}{3}}=\frac{3}{2}}$$

So the line we are looking for has a slope of $${\displaystyle \frac{3}{2}}$$.

Now determine the $y$-axis intercept of the line $h$ through $P(4|6)$ by substituting the point $P$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+b$

Plug in the values.

$${\displaystyle 6=\frac{3}{2}\cdot 4+b}$$ $|-6$

Simplify and subtract 6.

$0=b\Rightarrow b=0$

So the line equation is $${\displaystyle h(x):y=\frac{3}{2}x}$$.

First determine the slope of the line $h$ perpendicular to $${\displaystyle g(x):y=\frac{1}{3}x-2}$$ , where being perpendicular implies

$$m_2={\displaystyle -\frac{1}{m_1}}$$

Plug in the known values.

$$m_2={\displaystyle -\frac{1}{\frac{1}{3}}=-3}$$

So the line we are looking for has a slope of $-3$.

Now determine the $y$-axis intercept of the line $h$ through $P(2|5)$ by substituting the point $P$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+b$

Plug in the values.

$5=-3\cdot 2+b$ $|+6$

Simplify and add 6.

$11=b\Rightarrow b=11$

So the line equation is $h(x):y=-3x+11$.