Question

1) What is the equation of a line that is perpendicular to 2x+y=−4 and passes through the point (2, −8)

Enter your equation in the box.

Question 2 is the picture!

Thanks in advance for the help!

Answer 1

1.

`Let\ k:y=m_1x+b_1\ and\ l:y=m_2x+b_2,\ then\\\\l\ \perp\ k\iff m_1m_2=-1.\\\\\text{We have}\ k:2x+y=-4\qquad subtract\ 2x\ from\ both\ sides\\\\k:y=-2x-4\to m_1=-2\\\\l:y=mx+b\\\\l\ \perp\ k\iff-2m=-1\qquad divide\ both\ sides\ by\ (-2)\\\\m=(1)/(2)\\\\l:y=(1)/(2)x+b\\\\\text{The line}\ l\ \text{passes throuth the point (2, -8).}\\\text{Susbtitute the coordinates of the point to the equation of the line }\ l:\\\\-8=(1)/(2)(2)+b\\\\-8=1+b\qquad subtract\ 1\ from\ both\ sides\\\\-9=b\to b=-9`

`\boxed{y=(1)/(2)x-9}`

2.

If ΔABC is a right triangle, then we can use Pythagorean theorem.

The formula of a length of a line segment AB:

`|AB|=√((x_B-x_A)^2+(Y_B-y_A)^2)`

A(1, -1), B(3, 2)

`|AB|=√((3-1)^2+(2-(-1))^2)=√(2^2+3^2)=√(4+9)=√(13)`

(1) C(0, 2)

`|AC|=√((0-1)^2+(2-(-1))^2)=√((-1)^2+3^2)=√(1+9)=√(10)\\\\|BC|=√((0-3)^2+(2-2)^2)=√((-3)^2+0^2)=\sqrt9=3`

`√(13) >√(10) > 3\\\\3^2+(√(10))^2\stackrel{?}{=} (√(13))^2\\\\L=9+10=19\\R=13\\L\\eq R\\Not\ a\ Right\ Triangle`

(2) C(3, -1)

`|AC|=√((3-1)^2+(-1-(-1))^2)=√(2^2+0^2)=\sqrt4=2\\\\|BC|=√((3-3)^2+(-1-2)^2)=√(0^2+(-3)^2)=\sqrt9=3`

`√(13) > 3 > 2\\\\3^2+2^2\stackrel{?}{=}(√(13))^2\\\\L=9+4=13\\R=13\\L=R\\Right\ Triangle`

(3) C(0, 4)

`|AC|=√((0-1)^2+(4-(-1))^2)=√((-1)^2+5^2)=√(1+25)=√(26)\\\\|BC|=√((0-3)^2+(4-2)^2)=√((-3)^2+2^2)=√(9+4)=√(13)`

`√(26) > √(13)=√(13)\\\\(√(13))^2+(√(13))^2\stackrel{?}{=}(√(26))^2\\\\L=13+13=26\\R=26\\Right\ Triangle`