15) Formula logic and Coordinate geometry Flashcards
Types of questions asked in Formula Logic?
Formula Logic Qs ask us to understand the impact of a HYPOTHETICAL CHANGE to A FORMULA
> similar to proportions qs
(1) By what factor has the formula’s result changed? (FACTOR CHANGE)
> keep track of the ORIGINAL formula’s value in order to understand the “what factor” impact
> Factor = New value of formula / Old value of formula
(2) What will the percent change be if “x” occurs? (PERCENT CHANGE)
> percent change in the new outcome of a formula compared to the original outcome of a formula
Can either solve algebraically or using smart numbers
> algebraically –> rewrite new value in terms of old variables
Ex: On a certain planet, the density D can be expressed by the equation: D = 1/(3k)^3, where k is the radius of the planet. If k is halved, the density D would change by what factor?
D2 / D1 = x
D2 = x * D1
Ans 8 —> pull out (1/2)^3 in the denominator
If the radius of a circle is decreased by 10 percent, what will be the percentage change in the area of the circle?
Area of circle: pi * r^2
R2 = 0.9*r
Percent change formula: (area 2 - area1)/area 1 * 100
= (0.81r^2 - r^2)/r^2 * 100
= -19%
Ans: 19 percent
Coordinate pairs (ordered pair)
(x, y)
Every point has a unique set of coordinates consisting of an x and y value
Labeling quadrants and what points in each quadrant tell you
Top right is quad 1
Count going counterclockwise:
Top left: Quad 2
Bottom left: Quad 3
Bottom right: Quad 4
Few things to note:
> Quadrants are indicative of SIGNS of the ordered pair
> Quad 1 and Quad 3: x and y have the SAME SIGN (xy > 0 or x/y > 0)
> Quad 2 and Quad 4: x and y have OPPOSITE signs (xy < 0 or x/y < 0)
Creating a line segment
Requires at least two points to create a line
A line segment has a finite length, while a line theoretically extends to infinity
When are two points in the same quadrant?
x coordinates have the same sign AND y coordinates have the same sign
product of x coordinates > 0
AND
product of y coordinates > 0
(but don’t need x and y to necessarily have the same sign)
Formula for calculating slope of a line
What can the value of slope be?
slope m = Rise / Run = (Y2 - Y1) / (X2 - X1)
Positive slope = upward sloping
Negative slope = downward sloping
Value of slope of a line can be:
> 0 (horizontal) -> y = #
> positive
> negative
> undefined (Vertical line has undefined slope)
Lines with positive slopes and quadrants
** Rule 1: All positively sloped lines MUST cross through Quadrants I and III (though it may or may not intersect II and IV)
Rule 2: If the x intercept of a positively sloped line is negative, then it’s y-intercept WILL BE POSITIVE AND the line will intersect Quadrant II
Rule 3: If the x-intercept of a positively sloped line is 0, then it’s y-intercept is also 0 AND the line only passes through Quadrants I and III
> x-intercept and y-intercept have OPPOSITE signs
Rule 4: If the x-intercept of a positively sloped line is positive, then its y-intercept WILL BE NEGATIVE AND the line will intersect Quadrant IV
> x-intercept and y-intercept have OPPOSITE signs
How to check these rules?
> shifting a positively sloped line
Lines with negative slopes and quadrants
***** Rule 1: All negatively sloped lines MUST cross through Quadrants II and IV (though it may or may not intersect I and III)
Rule 2: If the x intercept of a negatively sloped line is negative, then it’s y-intercept WILL ALSO be negative AND the line will intersect Quadrant III
Rule 3: If the x-intercept of a negatively sloped line is 0, then it’s y-intercept is also 0 AND the line only passes through Quadrants II and IV
> All negatively sloped lines have an x-intercept and y-intercept WITH THE SAME SIGN
Rule 4: If the x-intercept of a negatively sloped line is positive, then its y-intercept WILL ALSO be positive AND the line will intersect Quadrant I
> All negatively sloped lines have an x-intercept and y-intercept WITH THE SAME SIGN
Lines with 0 slope
Horizontal lines - not mandatory quadrants it has to intersect
Rule 1: If y intercept of a horizontal line is POSITIVE, then the line intersects Quadrants I and II
Rule 2: If y intercept of a horizontal line is zero, then the line is the x-axis and does not pass through any of the quadrants
> only time the horizontal line intersects with the x axis
Rule 3: If y intercept of a horizontal line is negative, then the line intersects quadrants III and IV
Horizontal lines also have points that all have the SAME Y COORDINATE
> helpful for DS when given two points with unknown coordinates
(a, b) and (c, d) —> b=d
Lines with undefined slope
Undefined lines (zero nominator) - not mandatory quadrants it has to intersect
Rule 1: If x intercept of a vertical line is POSITIVE, then the line intersects Quadrants I and IV
Rule 2: If x intercept of a vertical line is zero, then the line is the y-axis and does not pass through any of the quadrants
> only time the vertical line intersects with the y axis
Rule 3: If x intercept of a vertical line is negative, then the line intersects quadrants II and III
Vertical lines also have points that all have the SAME X COORDINATE
Steepness of slope
Compare ABSOLUTE Value of slope
> The larger the absolute value of the slope of a line = the STEEPER the line
e.g. m = -3 —> read as “one over, 3 down” (NOT 3 left, one up)
Slope-intercept equation
y=mx+b
y = y coordinate for a point on the line
m = slope
x = corresponding x coordinate for a point on the line
b = y intercept of the line –> when x = 0
x intercept –> (x, 0)
Set y equal to zero to calculate x intercept
> just need to know (1) Slope of the line (2) y intercept
Working with the slope-intercept equation:
> need to REARRANGE and ISOLATE variable y in order to DETERMINE slope of the line and y intercept (if equation is presented in general form)
e.g., 3x + 5y = 8
What data is SUFFICIENT to fixate a line?
Aka to create an equation for a line
(1) two points
> can calc slope
> can calc y and x intercept
(2) Slope and one point
> incl. slope of a PARALLEL or perpendicular line
If you just have one point, NS (can rotate line about that point)
Equation for horizontal and vertical lines?
Equation for horizontal line is y = b —-> y intercept is b
> recognize special horizontal line is the x axis, y = 0
Equation for vertical line is x = a —-> x intercept is a
> recognize special vertical line is y axis, x = 0
Checking if a point is on a line
Sub in the value of x and y of the point INTO the equation of the line —> if it WORKS, then Yes, the point is on the line
Therefore, if some point A is on some line L, the x and y coordinates of point A MUST KEEP THE EQUATION OF LINE L in equality
for DS –> just need to see if you can plug in point into a full y=mx+b equation
What is the “Standard form” or “general form” equation of a line
Ax + By = C, where A, B and C are all constants
> best to convert the equation to slope-intercept point
Perpendicular lines
Product of slopes = -1
Implication:
> Slopes are NEGATIVE RECIPROCALS
e.g., 1 and -1
e.g., 2 and -1/2
e.g., 1/4 and -4
> Lines cross at one point and meet at a RIGHT ANGLE (creates 4 right angles)
Parallel lines
Lines with SAME SLOPE but DIFFERENT x and y intercepts
Implication:
> Parallel lines NEVER INTERSECT
Reflections of a point, a figure, a line, or a line segment OVER a line or a point:
Reflection of a POINT (x, y) over x axis
Same x coordinate, flip sign of y coordinate
(x, y) —> (x, -y)
e.g., (2, 3) —> (2, -3)
Implication:
> point and its reflection point are the SAME DISTANCE from the object being reflected over
> aka, the MIDPOINT between (x,y) and its reflection is on the x axis, y = 0
Notation: point A and A prime (A’)
ALSO: broader formula (y=b)
(x, y) —> (x, 2b - y)
Reflections of a point, a figure, a line, or a line segment OVER a line or a point:
Reflection of a POINT (x, y) over y axis
Same y coordinate, flip sign of x coordinate
(x, y) —> (-x, y)
e.g., (2, 3) –> (-2, 3)
Implication:
> point and its reflection point are the SAME DISTANCE from the object being reflected over
> aka, the MIDPOINT between (x,y) and its reflection is on the y axis, x = 0
Notation: point A and A prime (A’)
ALSO: broader formula (x=a)
(x, y) —> (2a - x, y)
Reflections of a point, a figure, a line, or a line segment OVER a line or a point:
Reflection of a POINT (x, y) over origin
Origin is a POINT –> change signs of BOTH x and y coordinate
(x, y) —> (-x, -y)
e.g., (2, 3) –> (-2, -3)
Implication:
> point and its reflection point are the SAME DISTANCE from the object being reflected over (origin is the midpoint)
Notation: point A and A prime (A’)
Reflections of a point, a figure, a line, or a line segment OVER a line or a point:
Reflection of a LINE SEGMENT AB over x axis
Reflect EACH of the ENDPOINTS of the line segment
Same thing applies to polygon like a triangle —> reflect each VERTEX then CONNECT the dots
Reflections of a point, a figure, a line, or a line segment OVER a line or a point:
Reflection over y = x
y = x —> line with slope 1 and y int = 0
(x, y) —> (y, x)
Flip x and y
Reflections of a point, a figure, a line, or a line segment OVER a line or a point:
Reflection over y = -x
y = -x —> line with slope -1 and y int = 0
(x, y) —> (-y, -x)
Flip x and y
Also change sign of each
Reflections of a point, a figure, a line, or a line segment OVER a line or a point:
Reflection over y = b
y = b —> horizontal line
(x, y) —-> (x, 2b - y)
Same x, y changes by 2b - y
e.g., point (5, -2) reflected over y = 1 becomes (5, 4)
OR VISUAL APPROACH –> equal distance from the reflection line
PAY ATTENTION TO x and y values
Reflections of a point, a figure, a line, or a line segment OVER a line or a point:
Reflection over x = a
x = a —> vertical line
(x, y) —> (2a - x, y)
Same y, x changes by 2a - x
e.g., point (5, -2) reflected over x = 2 becomes (-1, -2)
OR VISUAL APPROACH –> equal distance from the reflection line
PAY ATTENTION TO x and y values
Distance between two points
Distance = sqrt[ change x^2 + change y^2]
= sqrt[ (x2-x1)^2 + (y2-y1)^2 ]
Derived from the pythagorean theorem for right triangles
> pay attention to x2 vs x1 and y2 vs y1 (order matters)
”+” (not like difference of squares)
HOWEVER, if two points share an “x” or “y” coordinate –> short cut
> ABSOLUTE difference between different coordinate
e.g., share x coordinate —> look at change in y coordinates
e.g., share y coordinate –> look at change in x coordinates
A square is inscribed in a circle that has a diameter with endpoints (3, 4) and (3, -2). What is the area of the square?
Geometry rules:
Square can be comprised of 2 isosceles triangles with sides following this ratio:
1x: 1x: x*sqrt(2)
So once you know x*sqrt(2) = distance, then you can solve for x^2 = area of square
Since the two points share x coordinate, the distance between the two points is | 4 - -2 | = 6
therefore x^2 = 18
Calculating area of right triangles in xy plane
Plot the three vertexes to identify which two points create the hypotenuse
Midpoint of a line segment formula
Trying to find (x, y) of MIDPOINT: find the average of each coordinate
x coordinate = (x1 + x2)/2
y coordinate = (y1 + y2)/2
Line segment k, with a slope of zero, has endpoints (a, b) and (c, d). If the midpoint of line segment k is (10, 2), what is the value of a+b?
(1) d = 2
(2) a = c-2
Coordinate geometry
> always draw a picture so you can see slopes and hidden info
> in this case, line segment k has 0 slope so it is HORIZONTAL —-> endpoints have the SAME Y value
b = d = 2 based on the midpoint
(a+c)/2 = 10
a = 20 - c
Therefore, a + b = a + 2
= 20 - c + 2 = 22 - c
So we need value of a or value of c
(1) NS –> we already know this
(2) a = c - 2 = 20 - c —-> 1 var can solve for c
2c = 22
c = 11
so a = 11
Sufficient
B
Equation for a circle on the coordinate plane
Center at point (a, b) and radius r
MEMORIZE:
(x - a)^2 + (y - b)^2 = r^2
Might not need to expand equation PS
Derived from DISTANCE formula between center point (a, b) and any point on the circumference of the circle (x, y) = radius
A circle with a center at the ORIGIN (0, 0) will have equation
x^2 + y^2 = r^2
Graphing inequalities and their solution set
e.g., y > 2x + 4
e.g., 2x + 4 < y
Important notes:
> all inequalities HAVE TO BE in slope-intercept form (must have y isolated on the LEFT side of the inequality)
e.g., 2x + 4 < y —-> y > 2x + 4 —-> looking for region above 2x+4
Graphing inequalities:
(1) make sure you express as slope-intercept form
(2) graph normal linear equation first
e.g., y = 2x+4
(3) Turn line into inequality using DASHED LINES (for < >) or SOLID LINES (for <= >=) and SHADE region of applicable points
> there’s an infinite number of points that satisfy this inequality
What does this mean?
y > 2x +4 tells us that for ANY value of x, the value of y in our solution set will always be greater than the value of 2x+4
Similarly, region y < 2x + 4 tells us that for any value of x, the value of y in our solution set will always be less than the value of 2x + 4
THEREFORE GOAL IS TO SATISFY THE INEQUALITY
In the xy-plane, region R is defined by the inequality: y > 6x + 8. Which of the following points could be a solution to the inequality?
(2, 0)
(4, 12)
(30, 40)
(10, 50)
(4, 33)
Concept:
> for any value of x, our solution set must be greater than 6x + 8
Plug in values of x into 6x + 8 and pick answer that has y value GREATER than “manufactured” 6x+8 value
Ans (4, 33)
6x+8 when x=4 equals 32. Manufactured y (32) is LOWER than 33 provided
In other words:
33 > 6*4 + 8 (Yes)
Dealing with multiple inequalities in coordinate plane
Remember where the solution set of an inequality lies:
“Less than” inequality –> solution set lies only below the line
“Less than or equal to” inequality –> solution set lies ON the line AND below the line
“Greater than” inequality –> solution set lies only above the line
“Greater than or equal to” inequality –> solution set lies ON the line AND ABOVE the line
Graphing absolute value functions
e.g., | x | + | y | = 5
Easiest thing to do is to PLOT THE X and Y INTERCEPTS first
e.g., x = 0, y = +/-5
y = 0, x = +/-5
