Problems in Plane Analytic Geometry: Problems with Solutions

Solution:
The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by the formula:
Distance = $\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$

Solution:
[tex]AB=\sqrt{(5-2)^2+(-3-1)^2}=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5[/tex]

Solution:
[tex]y=\frac{y_2-y_1}{x_2-x_1}\rightarrow y=\frac{7-(-3)}{0-5}=\frac{10}{-5}=-2[/tex]

Solution:
The slope of the line is given by the formula $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ where $P_{1}(1,3)=(x_{1},y_{1})\qquad P_{2}(-2,1)=(x_{2},y_{2})$
so $m=\frac{1-3}{-2-1}=\frac{2}{3}$

The equation of the line is $y-y_{1}=m(x-x_{1})$
where $x_{1}=1$ and $y_{1}=3$ are the coordinates of $P_1$
$\Longrightarrow y-3=\frac{2}{3}(x-1)$
$y=\frac{2}{3}x-\frac{2}{3}+3=\frac{2}{3}x+\frac{7}{3}$

Answer: $y=\frac{2}{3}x+\frac{7}{3}$ is the equation of the line.

Solution:
$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-2}{2+2}=-\frac{1}{2}$

$b=\frac{x_{2}y_{1}-x_{1}y_{2}}{x_{2}-x_{1}}=\frac{2(2)-(-2)0}{2+2}=1$

then $y=-\frac{1}{2}x+1$

Solution:
The slope of a line which is parallel to x-axis is zero. So
[tex]y+1=0(x-4)\rightarrow y+1=0\rightarrow y=-1[/tex]

Solution:
Notice that the line passes through (0,-4), (2,0).
Therefore:
[tex]\frac{x}{2}+\frac{y}{-4}=1\rightarrow 2x-y=4[/tex]

Solution:
The distance between point $A(x_1, y_1)$ and
line $L: Ax + By + C = 0$ is given by the formula:
$distance=\frac{Ax_1+By_1+C}{A^2+B^2}$

[tex]d=\frac{|3\times 2-4\times (-3)+2|}{\sqrt{3^2+(-4)^2}}=\frac{|6+12+2|}{\sqrt{9+16}}=\frac{|20|}{\sqrt{25}}=\frac{20}{5}=4[/tex]

Solution:
[tex]A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\rightarrow S=\frac{1}{2}|0\times(0-3)+5\times(3-(-3))+0\times((-3)-0)|= \frac{30}{2}=15[/tex]

Solution:
Area is given by the formula $\pm \frac{1}{2}\det \left\vert \begin{array}{ccc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} \right\vert $

Area $=\pm \frac{1}{2}\det \left\vert \begin{array}{ccc} 1 & 1 & 1 \\ -1 & 2 & 1 \\ 2 & -1 & 1 \end{array} \right\vert =\frac{3}{2}$

Solution:
[tex]x=-4 \rightarrow x+4=0 \rightarrow A=1, B=0, C=4[/tex]
[tex]7=\frac{|1\times(2a-3)+0\times 5+4|}{\sqrt{1^2+0^2}}=\frac{|2a-3+4|}{\sqrt{1}} =\frac{|2a+1|}{1}=|2a+1|\rightarrow[/tex]
[tex]|2a+1|=7 \rightarrow \begin{cases} 2a+1=7 \\ 2a+1=-7 \end{cases} \rightarrow \begin{cases} 2a=6 \\ 2a=-8 \end{cases}\rightarrow \begin{cases} a=3 \\ a=-4 \end{cases}[/tex]

Solution:
[tex]D_1:x-y=1[/tex]
[tex]D_2=2x+y=5[/tex]
[tex]\rightarrow 3x=6 \rightarrow x=2, y=1[/tex]
[tex](2m-5)x-my=3\rightarrow (2m-5)2-m=3 \rightarrow 4m-10-m=3 \rightarrow 3m=13 \rightarrow m=\frac{13}{3}[/tex]