91Ó°ÊÓ

The given figure is:

We have to find the slope of $\stackrel{↔}{AB}$.

Slope formula with two points$\left(x_1,y_1\right)\text{ â¶Ä‰a²Ô»å â¶Ä‰}\left(x_2,y_2\right)$is:

$m=\frac{y_2−y_1}{x_2−x_1}$

The given points are$A\left(1,0\right)$and$B\left(3,3\right)$.

$m=\frac{3−0}{\left(3\right)−\left(1\right)}=\frac{3}{2}=1.5$

So, the slope of$\stackrel{↔}{AB}$ is 1:5

The given figure is:

We have to find $\tan n^{∘}$.

Slope formula with two points$\left(x_1,y_1\right)\text{ â¶Ä‰a²Ô»å â¶Ä‰}\left(x_2,y_2\right)$is:

$\tan \mathrm{θ}=m=\frac{y_2−y_1}{x_2−x_1}$

Here$\mathrm{θ}$ = the angle that the lines makes with the positive direction of the$x−\text{axis}$ .

The given points$A\left(1,0\right)$are and$B\left(3,3\right)$.

$\begin{array}{l}\tan n^{∘}=m=\frac{3−0}{\left(3\right)−\left(1\right)}=\frac{3}{2}=1.5\\ \tan n^{∘}=1.5\end{array}$

So, the value of$\tan n^{∘}$ is 1.5.

The given figure is:

We have to determine whether the statement is true or false and explain.

Slope formula with two points$\left(x_1,y_1\right)\text{ â¶Ä‰a²Ô»å â¶Ä‰}\left(x_2,y_2\right)$is:

$m=\frac{y_2−y_1}{x_2−x_1}$

The given points are$A\left(1,0\right)$and$B\left(3,3\right)$.

$m=\frac{3−0}{\left(3\right)−\left(1\right)}=\frac{3}{2}=1.5$

Also,

$\tan n^{∘}=\frac{\text{Perpendicular}}{\text{Base}}=\frac{3}{2}=1.5$

So, the given statement is true.