Question

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

Answer 1

1.

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

$\boxed{y=\dfrac{1}{2}x-9}$

2.

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

The formula of a length of a line segment AB:

$|AB|=\sqrt{(x_B-x_A)^2+(Y_B-y_A)^2}$

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

$|AB|=\sqrt{(3-1)^2+(2-(-1))^2}=\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13}$

(1) C(0, 2)

$|AC|=\sqrt{(0-1)^2+(2-(-1))^2}=\sqrt{(-1)^2+3^2}=\sqrt{1+9}=\sqrt{10}\\\\|BC|=\sqrt{(0-3)^2+(2-2)^2}=\sqrt{(-3)^2+0^2}=\sqrt9=3$

$\sqrt{13} >\sqrt{10} > 3\\\\3^2+(\sqrt{10})^2\stackrel{?}{=} (\sqrt{13})^2\\\\L=9+10=19\\R=13\\L\neq R\\Not\ a\ Right\ Triangle$

(2) C(3, -1)

$|AC|=\sqrt{(3-1)^2+(-1-(-1))^2}=\sqrt{2^2+0^2}=\sqrt4=2\\\\|BC|=\sqrt{(3-3)^2+(-1-2)^2}=\sqrt{0^2+(-3)^2}=\sqrt9=3$

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

(3) C(0, 4)

$|AC|=\sqrt{(0-1)^2+(4-(-1))^2}=\sqrt{(-1)^2+5^2}=\sqrt{1+25}=\sqrt{26}\\\\|BC|=\sqrt{(0-3)^2+(4-2)^2}=\sqrt{(-3)^2+2^2}=\sqrt{9+4}=\sqrt{13}$

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