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

${\displaystyle m_2=-\frac{1}{m_1}}\backslash displaystyle\; m\_2=-\backslash frac1\{m\_1\}$

Plug in the known values.

${\displaystyle m_2=-\frac{1}{3}}\backslash displaystyle\; m\_2=-\backslash frac1\{3\}$

So the line we are looking for has a slope of ${\displaystyle -\frac{1}{3}}\backslash displaystyle-\backslash frac13$.

Now determine the $yy$-axis intercept of the line $hh$ through $P(3\mathrm{\mid }5)P(3|5)$ by substituting the point $PP$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+bh(x\_p):y\_p=m\_2x\_p+b$

Plug in the values.

$5=-\frac{1}{3}\cdot 3+b\mid +15=-\backslash frac13\backslash cdot3+b\backslash ;\backslash ;\backslash ;\backslash left|+1\backslash right.$

Simplify and add $11$.

$6=b\iff b=66=b\backslash Leftrightarrow\; b=6$

So the line equation is ${\displaystyle h(x):y=-\frac{1}{3}x+6}\backslash displaystyle\; h(x):y=-\backslash frac13x+6$.

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

$m_2={\displaystyle -\frac{1}{m_1}}m\_2=\backslash displaystyle\; -\backslash frac1\{m\_1\}$

Plug in the known values.

${\displaystyle m_2=-\frac{1}{0.5}=-\frac{1}{\frac{1}{2}}=-2}\backslash displaystyle\; m\_2=-\backslash frac1\{0.5\}=-\backslash frac1\{\backslash frac12\}=-2$

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

Now determine the $yy$-axis intercept of the line $hh$ through $P(1\mathrm{\mid }2)P(1|2)$ by substituting the point $PP$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+bh(x\_p):y\_p=m\_2x\_p+b$

Plug in the values.

$2=-2\cdot 1+b2=-2\backslash cdot1+b$ $\mathrm{\mid }+2|+2$

Simplify and add 2.

$4=b\Rightarrow b=44=b\backslash Rightarrow\; b=4$

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

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

${\displaystyle m_2=-\frac{1}{m_1}}\backslash displaystyle\; m\_2=-\backslash frac1\{m\_1\}$

Plug in the known values.

${\displaystyle m_2=-\frac{1}{-5}=0.2}\backslash displaystyle\; m\_2=-\backslash frac1\{-5\}=0.2$

So the line we are looking for has a slope of $0.20.2$.

Now determine the $yy$-axis intercept of the line $hh$ through $P(-10\mathrm{\mid }1)P(-10|1)$ by substituting the point $PP$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+bh(x\_p):y\_p=m\_2x\_p+b$

Plug in the values.

$1=0.2\cdot (-10)+b1=0.2\backslash cdot(-10)+b$ $\mathrm{\mid }+2|+2$

Simplify and add 2.

$3=b\Rightarrow b=33=b\backslash Rightarrow\; b=3$

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

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

${\displaystyle m_2=-\frac{1}{m_1}}\backslash displaystyle\; m\_2=-\backslash frac1\{m\_1\}$

Plug in the known values.

$m_2=-{\displaystyle \frac{1}{4}=-0.25}m\_2=-\backslash displaystyle\backslash frac14=-0.25$

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

Now determine the $yy$-axis intercept of the line $hh$ through $P(2\mathrm{\mid }-5)P(2|-5)$ by substituting the point $PP$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+bh(x\_p):y\_p=m\_2x\_p+b$

Plug in the values.

$-5=-0.25\cdot 2+b-5=-0.25\backslash cdot2+b$ $+0.5+0.5$

Simplify and add $0.50.5$.

$-4.5=b\Rightarrow b=-4.5-4.5=b\backslash Rightarrow\; b=-4.5$

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

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

${\displaystyle m_2=-\frac{1}{m_1}}\backslash displaystyle\; m\_2=-\backslash frac1\{m\_1\}$

Plug in the known values.

${\displaystyle m_2=-\frac{1}{-\frac{2}{3}}=\frac{3}{2}}\backslash displaystyle\; m\_2=-\backslash frac1\{-\backslash frac23\}=\backslash frac32$

So the line we are looking for has a slope of ${\displaystyle \frac{3}{2}}\backslash displaystyle\backslash frac32$.

Now determine the $yy$-axis intercept of the line $hh$ through $P(4\mathrm{\mid }6)P(4|6)$ by substituting the point $PP$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+bh(x\_p):y\_p=m\_2x\_p+b$

Plug in the values.

${\displaystyle 6=\frac{3}{2}\cdot 4+b}\backslash displaystyle6=\backslash frac32\backslash cdot4+b$ $\mathrm{\mid }-6|-6$

Simplify and subtract 6.

$0=b\Rightarrow b=00=b\backslash Rightarrow\; b=0$

So the line equation is ${\displaystyle h(x):y=\frac{3}{2}x}\backslash displaystyle\; h(x):y=\backslash frac32x$.

First determine the slope of the line $hh$ perpendicular to ${\displaystyle g(x):y=\frac{1}{3}x-2}\backslash displaystyle\; g(x):y=\backslash frac13x-2$ , where being perpendicular implies

$m_2={\displaystyle -\frac{1}{m_1}}m\_2=\backslash displaystyle-\backslash frac1\{m\_1\}$

Plug in the known values.

$m_2={\displaystyle -\frac{1}{\frac{1}{3}}=-3}m\_2=\backslash displaystyle\; -\backslash frac1\{\backslash frac13\}=-3$

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

Now determine the $yy$-axis intercept of the line $hh$ through $P(2\mathrm{\mid }5)P(2|5)$ by substituting the point $PP$ into the general line equation.

$h(x_p):y_p=m_2{x}_p+bh(x\_p):y\_p=m\_2x\_p+b$

Plug in the values.

$5=-3\cdot 2+b5=-3\backslash cdot2+b$ $\mathrm{\mid }+6|+6$

Simplify and add 6.

$11=b\Rightarrow b=1111=b\backslash Rightarrow\; b=11$

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