I need help with finding line equations, showing tri. prop's

[Cinnamon]

Here is the problem i need help with:

Find the equation of the lines satisfying the given conditions parallel to the x-axis and passes through the point (-5, -3)

And this one as well:

Use slope to show that Triangle ABC with vertices a(0, 2), b(2, 3), and c (1, 5) is a right triangle.

Please help quickly. Thank you!

 

[soroban]

Re: I NEED HELP WITH THESE MATH PROBLEMS

Hello, Cinnamon!

Find the equation of the line parallel to the x-axis and passes through the point (-5, -3)

"Parallel to the x-axis" means the line is horizontal; its slope is $\displaystyle m\,=\,0$

Point-slope Formula: the line through $\displaystyle (x_1,y_1)$ with slope $\displaystyle m$ is:
$\displaystyle \;\;\;y\,-\,y_1\;=\;m(x\,-\,x_1)$

Our line has the point $\displaystyle (-5,3)$ and $\displaystyle m\,=\,0$

Its equation is: $\displaystyle \,y\,-\,3\;=\;0(x\,-\,[-5])\;\;\Rightarrow\;\;y\,=\,3$

Use slope to show that Triangle ABC with vertices A(0,2), B(2,3), and C (1,5) is a right triangle.

The slope of AB is: $\displaystyle \,m_{AB} \:=\:\frac{3\,-\,2}{2\,-\,0}\:=\:\frac{1}{2}$

The slope of BC is: $\displaystyle \,m_{BC}\:=\:\frac{5\,-\,3}{1\,-\,2}\:=\:-2$

Since these slopes are negative reciprocals of each other, $\displaystyle AB\,\perp\,BC$

Therefore: $\displaystyle \,\angle ABC\,=\,90^o\;\;\Rightarrow\;\;\Delta ABC$ is a right triangle.

 

[jacket81]

Soroban just made a minor mistake with first problem.
It should be y=-3, not y=3.
He had the point down wrong.