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Coordinate geometry - Line equation

For COMPETITION
Number of Total Problems: 6.
FOR PRINT ::: (Book)

Problem Num : 1

From : NCTM

Type: None

Section:Coordinate geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Line equation

Analysis

Solution/Answer

Problem Num : 2

From : NCTM

Type: None

Section:Coordinate geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Line equation

Analysis

Solution/Answer

Problem Num : 3

From : AMC10B

Type:

Section:Coordinate geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Define $aclubsuit b=a^2b-ab^2$ . Which of the following describes the set of points $(x, y)$ for which $xclubsuit y=yclubsuit x$ ?
$extbf{(A)} ext{a finite set of points}\ qquad extbf{(B)} ext{one line}\ qquad extbf{(C)} ext{two parallel li...$
'

Category Line equation

Analysis

Solution/Answer
$xclubsuit y = x^2y-xy^2$ and $yclubsuit x = y^2x-yx^2$ . Therefore, we have the equation $x^2y-xy^2 = y^2x-yx^2$ Factoring out a $-1$ gives $x^2y-xy^2 = -(x^2y-xy^2)$ Factoring both sides further, $xy(x-y) = -xy(x-y)$ . It follows that if $x=0$ , $y=0$ , or $(x-y)=0$ , both sides of the equation equal 0. By this, there are 3 lines ( $x=0$ , $y=0$ , or $x=y$ ) so the answer is $oxed{ extbf{(E)} ext{ three lines}}$ .

Answer:

Problem Num : 4

From : AMC10B

Type:

Section:Coordinate geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$ . Find the length of segment $AE$ .

$mathrm{(A)} frac{4sqrt{5}}{3} qquadmathrm{(B)} frac{5sqrt{5}}{3} qquadmathrm{(C)} frac{12sqrt{5}}{7} qquadmat...$
'

Category Line equation

Analysis

Solution/Answer
Solution 1

Let $l_1$ be the line containing $A$ and $B$ and let $l_2$ be the line containing $C$ and $D$ . If we set the bottom left point at $(0,0)$ , then $A=(0,3)$ , $B=(6,0)$ , $C=(4,2)$ , and $D=(2,0)$ .
The line $l_1$ is given by the equation $y=m_1x+b_1$ . The $y$ -intercept is $A=(0,3)$ , so $b_1=3$ . We are given two points on $l_1$ , hence we can compute the slope, $m_1$ to be $frac{0-3}{6-0}=frac{-1}{2}$ , so $l_1$ is the line $y=frac{-1}{2}x+3$
Similarly, $l_2$ is given by $y=m_2x+b_2$ . The slope in this case is $frac{2-0}{4-2}=1$ , so $y=x+b_2$ . Plugging in the point $(2,0)$ gives us $b_2=-2$ , so $l_2$ is the line $y=x-2$ .
At $E$ , the intersection point, both of the equations must be true, so $egin{align*}y=x-2, y=frac{-1}{2}x+3 &Rightarrow x-2=frac{-1}{2}x+3 \&Rightarrow x=frac{10}{3} \&Rightar...$
We have the coordinates of $A$ and $E$ , so we can use the distance formula here: $sqrt{left(frac{10}{3}-0 ight)^2+left(frac{4}{3}-3 ight)^2}=frac{5sqrt{5}}{3}$
which is answer choice $oxed{ ext{B}}$

Solution 2

Draw the perpendiculars from $A$ and $B$ to $CD$ , respectively. As it turns out, $BC perp CD$ . Let $F$ be the point on $CD$ for which $AFperp CD$ .
$mangle AFE=mangle BCE=90^circ$ , and $mangle AEF=mangle CEB$ , so by AA similarity, $riangle AFEsim riangle BCE Rightarrow frac{AF}{AE}=frac{BC}{BE}$
By the Pythagorean Theorem, we have $AB=sqrt{3^2+6^2}=3sqrt{5}$ , $AF=sqrt{2.5^2+2.5^2}=2.5sqrt{2}$ , and $BC=sqrt{2^2+2^2}+2sqrt{2}$ . Let $AE=x$ , so $BE=3sqrt{5}-x$ , then $frac{2.5sqrt{2}}{x}=frac{2sqrt{2}}{3sqrt{5}-x}$ $x=frac{5sqrt{5}}{3}$
This is answer choice $oxed{ ext{B}}$
Also, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.

Answer:

Problem Num : 5

From : AMC10B

Type:

Section:Coordinate geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A line with slope $3$ intersects a line with slope $5$ at point $(10,15)$ . What is the distance between the $x$ -intercepts of these two lines?
$extbf{(A) } 2 qquad extbf{(B) } 5 qquad extbf{(C) } 7 qquad extbf{(D) } 12 qquad extbf{(E) } 20$
'

Category Line equation

Analysis

Solution/Answer
Using the point-slope formula, the equation of each line is
$y-15=3(x-10) longrightarrow y=3x-15\y-15=5(x-10) longrightarrow y=5x-35$
Substitute in $y=0$ to find the $x$ -intercepts.
$0=3x-15longrightarrowx=50=5x-35longrightarrowx=7$ The difference between them is $7-5=oxed{ extbf{(A) } 2}$ .

Answer:

Problem Num : 6

From : AMC10B

Type:

Section:Coordinate geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
In rectangle $ABCD$ , we have $A=(6,-22)$ , $B=(2006,178)$ , $D=(8,y)$ , for some integer $y$ . What is the area of rectangle $ABCD$ ?
$mathrm{(A) } 4000qquad mathrm{(B) } 4040qquad mathrm{(C) } 4400qquad mathrm{(D) } 40,000qquad mathrm{(E) }...$
'

Category Line equation

Analysis

Solution/Answer
Solution 1

Let the slope of $AB$ be $m_1$ and the slope of $AD$ be $m_2$ .
$m_1 = frac{178-(-22)}{2006-6} = frac{1}{10}$
$m_2 = frac{y-(-22)}{8-6} = frac{y+22}{2}$
Since $AB$ and $AD$ form a right angle:
$m_2 = -frac{1}{m_1}$
$m_2 = -10$
$frac{y+22}{2} = -10$
$y = -42$
Using the distance formula:
$AB = sqrt{ (2006-6)^2 + (178-(-22))^2 } = sqrt{ (2000)^2 + (200)^2 } = 200sqrt{101}$
$AD = sqrt{ (8-6)^2 + (-42-(-22))^2 } = sqrt{ (2)^2 + (-20)^2 } = 2sqrt{101}$
Therefore the area of rectangle $ABCD$ is $200sqrt{101}cdot2sqrt{101} = 40,400 Rightarrow E$

Solution 2

This solution is the same as Solution 1 up to the point where we find that $y=-42$ .
We build right triangles so we can use the Pythagorean Theorem. The triangle with hypotenuse $AB$ has legs $200$ and $2000$ , while the triangle with hypotenuse $AD$ has legs $2$ and $20$ . Aha! The two triangles are similar, with one triangle having side lengths $100$ times the other!
Let $AD=x$ . Then from our reasoning above, we have $AB=100x$ . Finally, the area of the rectangle is $100x(x)=100x^2=100(20^2+2^2)=100(400+4)=100(404)=oxed{40400 ext{ (E)}}$ .

Answer:

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