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Practice
Problems in Plane Analytic Geometry
Easy
Normal
Problems in Plane Analytic Geometry: Problems with Solutions
Problem 1
Find the distance between A(5, -3) and B(2, 1).
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]
Problem 2
Find the slope of a line, which passes through point А(5, -3) and meets y axis at 7.
Solution:
[tex]y=\frac{y_2-y_1}{x_2-x_1}\rightarrow y=\frac{7-(-3)}{0-5}=\frac{10}{-5}=-2[/tex]
Problem 3
The equation of the line through points $P_{1}(1,3)$ and $P_{2}(-2,1)$ is
$y=\frac{2}{3}x-\frac{7}{3}$
$y=\frac{3}{2}x+\frac{7}{3}$
$y=\frac{2}{3}x+\frac{7}{3}$
$y=\frac{3}{2}x+\frac{3}{7}$
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.
Problem 4
A line with equation $y=mx+b$ passes through the points $P_{1}(-2,2)$ and $P_{2}(2,0)$. What are the values of $m$ and $b$?
$m=-\frac{1}{2}\qquad b=1$
$m=\frac{1}{2}\qquad b=1$
$m=1\qquad b=\frac{1}{2}$
$m=2\qquad b=0$
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 $y=-\frac{1}{2}x+1$
Problem 5
Find the equation of a line which passes through A(4, -1) and is parallel to
x
axis.
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]
Problem 6
Find the equation of a line which intercept
y
axis at -4 and also intercept
x
axis at 2.
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]
Problem 7
Find the distance between A(2, -3) and the line L:
3x - 4y + 2 = 0
.
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]
Problem 8
Find the area of a triangle with vertices: A(0, -3), B(5, 0), C(0, 3).
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]
Problem 9
What is the area of triangle with vertices $P(1,1),Q(-1,2),R(2,-1)$
$\frac{3}{2}$
$\frac{5}{2}$
3
6
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 $=\pm \frac{1}{2}\det \left\vert \begin{array}{ccc} 1 & 1 & 1 \\ -1 & 2 & 1 \\ 2 & -1 & 1 \end{array} \right\vert =\frac{3}{2}$
Problem 10
If the distance of A(2x - 3, 5) from line x = -4 is equal to 7, then find the value of
a
.
Answer format:
m,n where m < n
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]
Problem 11
Find the value of
m
, such that D
1
, D
2
, D
3
meet each other at one point.
D
1
: x - y = 1
D
2
: 2x + y = 5
D
3
: (2m - 5)x - my = 3
Answer format:
p/q
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]
Easy
Normal
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