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06 May 2023, 12:40
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Show SpoilerDOWNLOAD PDF: How To Solve: Co-ordinate Geometry Basics
Attachment:
How To Solve: Co-ordinate Geometry Basics
Attached pdf of this Article as SPOILER at the top! Happy learning!
Hi All,
I have recently uploaded a video on YouTube to discuss Co-ordinate Geometry Basics in Detail:
Following is covered in the video
- ¤ XY or 2D Plane
¤ Introduction to Quadrants
¤ x-intercept and y-intercept
¤ Distance between two points
¤ Distance of a point from origin
¤ Slope of a line
¤ Sign of slope of a line
¤ Slope of Parallel and Perpendicular lines
XY or 2D Plane
XY Plane is a 2D Plane which contains the x-axis and the y-axis. x-axis is horizontal axis and y-axis is vertical axis.
¤ Each point in the XY Plane has two co-ordinates (x,y)
¤ x is the x co-ordinate, y is the y co-ordinate
Attachment:
xy plane.jpg [ 13.63 KiB | Viewed 2354 times ]
Introduction to Quadrants
x-axis and y-axis divide the XY plane into 4 quadrants.
¤ \(1^{st}\) Quadrant : x +ve, y +ve
¤ \(2^{nd}\) Quadrant : x -ve, y +ve
¤ \(3^{rd}\) Quadrant : x -ve, y -ve
¤ \(4^{th}\) Quadrant : x +ve, y -ve
Attachment:
Quadrants.jpg [ 31.7 KiB | Viewed 2326 times ]
x-intercept and y-intercept
x-intercept : Point where a line touches the x-axis
y-intercept : Point where a line touches the y-axis
Attachment:
Intercepts.jpg [ 19.33 KiB | Viewed 2336 times ]
Distance between two points
Distance, d, between two points A (x1,y1) and B(x2,y2) on a XY plane is given by
d = \(\sqrt{((x_2 − x_1)^2 + (y_2 − y_1)^2)}\)
Attachment:
Distance between points.jpg [ 18.78 KiB | Viewed 2364 times ]
Distance of a point from Origin
Distance, d, of a point A (a,b) from origin (0,0) is given by
d = \(\sqrt{((a − 0)^2 + (b − 0)^2)}\) = \(\sqrt{a^2 + b^2}\)
Attachment:
Distance from origin.jpg [ 17.25 KiB | Viewed 2347 times ]
Slope of a Line
Slope of a line is an indication of how inclined the line is as compared to positive x-axis.
¤ Slope(m) of line(l) passing through points A and B is given by
m = \(\frac{(y_2 − y_1)}{(x_2 − x_1)}\)
Attachment:
Slope of a line.jpg [ 18.51 KiB | Viewed 2331 times ]
Sign of slope of a line
¤ Positive Slope: Line tilted towards right
¤ Negative Slope: Line tilted towards left
¤ Zero Slope: Line parallel to x-axis
¤ Infinite Slope: Line parallel to y-axis
Attachment:
Sign of slope.jpg [ 22.08 KiB | Viewed 2343 times ]
Slope of Parallel and Perpendicular Lines
If two lines are Parallel, then their slopes will be equal.
¤ If we have two parallel lines \(L_1\) and \(L_2\), with below equations
¤ Line L1 : y = \(m_1\)x + \(c_1\)
¤ Line L2 : y = \(m_2\)x + \(c_2\)
¤ then \(m_1\) = \(m_2\)
Attachment:
Parallel Lines.jpg [ 4.93 KiB | Viewed 2365 times ]
If two lines are Perpendicular, then product of their slopes will be equal to -1
¤ If we have two perpendicular lines \(L_1\) and \(L_2\), with below equations
¤ Line L1 : y = \(m_1\)x + \(c_1\)
¤ Line L2 : y = \(m_2\)x + \(c_2\)
¤ then \(m_1\) * \(m_2\) = -1
Attachment:
Perpendicular Lines.jpg [ 3.41 KiB | Viewed 2317 times ]
Link to next post on Equation of a Line is here
Watch the following video to learn about Equation of a Line
Hope it helps!
Good Luck!
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12 May 2023, 20:34
Kudos
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Show SpoilerDOWNLOAD PDF: How To Solve: Equation of a Line
Attachment:
How To Solve: Equation of a Line
Attached pdf of this Article as SPOILER at the top! Happy learning!
Hi All,
I have recently uploaded a video on YouTube to discuss Equation of a Line in Detail:
Following is covered in the video
- ¤ Equation of a Line: Two Point Form
¤ Equation of a Line: Point and Slope Form
¤ Equation of a Line: Intercept Form
¤ Generic Equation of a line (Point and Intercept Form)
¤ Equation of Horizontal and Vertical Lines
Equation of a Line: Two Point Form
Let's say we have a line passing through two point A(\(x_1\),\(y_1\)) and B(\(x_2\),\(y_2\)).
Let's take a point C (x,y) on the line and between A and B as shown below.
Attachment:
EL img-1.jpg [ 20.58 KiB | Viewed 2014 times ]
Slope of the Part CA of the line = Slope of the part BA of the line = Slope of the line
=> \(\frac{y - y_1}{x - x_1}\) = \(\frac{y_2 - y_1}{x_2 - x_1}\) [ Watch this video if you want to know about the slope of the line ]
=> Equation of a line in Two Point Form as y - \(y_1\) = \(\frac{y_2 - y_1}{x_2 - x_1}\) * (x - \(x_1\))
where x and y are variables and value of (\(x_1\),\(y_1\)) and (\(x_2\),\(y_2\)) will be given to us in the problem.
Equation of a Line: Point and Slope Form
Let's take the same line which was passing through point A and B and has a slope m
Attachment:
EL img-1.jpg [ 20.58 KiB | Viewed 2014 times ]
Slope, m = \(\frac{y_2 - y_1}{x_2 - x_1}\)
Using, y - \(y_1\) = \(\frac{y_2 - y_1}{x_2 - x_1}\) * (x - \(x_1\))
Equation of a line in Point and Slope Form as y - \(y_1\) = m * (x - \(x_1\))
Equation of a Line: Intercept Form
Let's say we have a line which intercepts X-Axis at point A(a,0) and Y-Axis at point B(0,b), as shown below.
Attachment:
EL img-2.jpg [ 20.17 KiB | Viewed 2012 times ]
Using, \(\frac{y - y_1}{x - x_1}\) = \(\frac{y_2 - y_1}{x_2 - x_1}\) we get
\(\frac{y - 0}{x - a}\) = \(\frac{b - 0}{0 - a}\)
=> ay = -bx + ab
=> bx + ay = ab
Dividing both the sides by ab we get
Equation of a line in Intercept Form as \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
Generic Equation of a line (Point and Intercept Form)
Let's say we have a line which intercepts Y-Axis at point B(0,b) and has a slope m, as shown below.
Attachment:
EL - generic equation.jpg [ 20.28 KiB | Viewed 2014 times ]
Using, y - \(y_1\) = m * (x - \(x_1\)) and substituting the value of Point B we get
y - b = m * (x - 0)
Generic Equation of a line (Point and Intercept Form) as y = mx + b
where m is the slope of the line and B is the y intercept.
Equation of horizontal and vertical lines
Let's say we have a line parallel to X-Axis and intersecting Y-Axis at point B(0,b) and
a line which is parallel to Y-Axis and intercepting the X-Axis at point A(a,0) as shown below
Attachment:
EL - Horizontal and parallel lines.jpg [ 21.02 KiB | Viewed 2009 times ]
Equation of Horizontal Line
Now, all the points on this line will be at the same distance b from X-Axis and will have the y-coordinate as b
=> Equation of Horizontal line will be y = b [constant]
Equation of Vertical Line
Now, all the points on this line will be at the same distance a from Y-Axis and will have the x-coordinate as a
=> Equation of Vertical line will be x = a [constant]
Watch the following video to learn the Basics of Co-ordinate Geometry
Hope it helps!
Good Luck!
15 May 2023, 09:17
Kudos
Bookmarks
Show SpoilerDOWNLOAD PDF: How To Solve: Slope of a Line
Attachment:
How To Solve: Slope of a Line
Attached pdf of this Article as SPOILER at the top! Happy learning!
Hi All,
I have recently uploaded a video on YouTube to discuss Slope of a Line in Detail:
Following is covered in the video
- ¤ Proof of Slope of a Line formula
¤ How to find slope of a Generic Line
¤ Sign of Slope of a Line
¤ Slope of Parallel and Perpendicular Lines
¤ Solved Problem : Find slope of a line given two points
¤ Solved Problem : Find slope of a line given equation of a line
¤ Solved Problem : Find value of a variable given the slope of a line
¤ Solved Problem : Find equation of a line parallel to a line and passing through a point
¤ Solved Problem : Find equation of a line perpendicular to a line and passing through a point
Proof of Slope of a Line formula
Let's say we have a line(l) passing through two point A(\(x_1\),\(y_1\)) and B(\(x_2\),\(y_2\)) and having a slope of m.
Lets draw perpendicular lines as shown in the below figure to complete △ABC
Let ∠BAC = θ
Attachment:
Slope of a Line - 1.jpg [ 16.96 KiB | Viewed 2416 times ]
Slope of a line, m = tan(Angle made by the line with positive X-Axis) = tan(θ) = Opposite Side / Adjacent Side = \(\frac{BC}{AC}\) = \(\frac{y_2 - y_1}{x_2 - x_1}\)
How to find slope of a Generic Line
Let's say we have a generic equation of the line given by ax + by = c
=> by = -ax + c
=> y = \(\frac{-a}{b}\)x + \(\frac{c}{b}\)
Comparing above equation with generic equation of a line y = mx + b we get
m = \(\frac{-a}{b}\)
=> Slope of the line ax + by = c, is given by m = \(\frac{-a}{b}\) = -Coefficient of x / Coefficient of y
Sign of Slope of a Line
¤ Positive Slope: Line tilted towards right
¤ Negative Slope: Line tilted towards left
¤ Zero Slope: Line parallel to x-axis
¤ Infinite Slope: Line parallel to y-axis
Attachment:
Sign of slope.jpg [ 22.08 KiB | Viewed 2385 times ]
Slope of Parallel and Perpendicular Lines
If two lines are Parallel, then their slopes will be equal.
¤ If we have two parallel lines \(L_1\) and \(L_2\), with below equations
¤ Line L1 : y = \(m_1\)x + \(c_1\)
¤ Line L2 : y = \(m_2\)x + \(c_2\)
¤ then \(m_1\) = \(m_2\)
Attachment:
Parallel Lines.jpg [ 4.93 KiB | Viewed 2381 times ]
If two lines are Perpendicular, then product of their slopes will be equal to -1
¤ If we have two perpendicular lines \(L_1\) and \(L_2\), with below equations
¤ Line L1 : y = \(m_1\)x + \(c_1\)
¤ Line L2 : y = \(m_2\)x + \(c_2\)
¤ then \(m_1\) * \(m_2\) = -1
Attachment:
Perpendicular Lines.jpg [ 3.41 KiB | Viewed 2385 times ]
Solved Problem : Find slope of a line given two points
Q1. Find the slope of the line passing through two points (1,3) and (3,5).
Sol 1: Slope of a line is given by m = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{5 - 3}{3 - 1}\)
= \(\frac{2}{2}\) = 1
= \(\frac{2}{2}\) = 1
Solved Problem : Find slope of a line given equation of a line
Q2. Find the slope of the line 2x + 3y = 5.
Sol 2: Slope of the line ax + by = c, is given by m = -Coefficient of x / Coefficient of y
= \(\frac{-2}{3}\)
= \(\frac{-2}{3}\)
Solved Problem : Find value of a variable given the slope of a line
Q3. A line with slope 2 passes through the points (x,5) and (2,x). Find the value of x.
Sol 3: Slope of a line is given by m = \(\frac{y_2 - y_1}{x_2 - x_1}\)
= \(\frac{x - 5}{2 - x}\) = 2 (given)
=> x - 5 = 4 - 2x
=> 3x = 9
=> x = \(\frac{9}{3}\) = 3
= \(\frac{x - 5}{2 - x}\) = 2 (given)
=> x - 5 = 4 - 2x
=> 3x = 9
=> x = \(\frac{9}{3}\) = 3
Solved Problem : Find equation of a line parallel to a line and passing through a point
Q4. Find the equation of the line which is parallel to the line 2x + 4y = 5 and passes through the point (2,3).
Sol 4: Equation of line 2x + 4y = 5 = \(\frac{-2}{4}\) = \(\frac{-1}{2}\)
=> Parallel line will have the same slope = \(\frac{-1}{2}\)
=> Equation of the line is given by y - \(y_1\) = m * (x - \(x_1\))
=> y - 3 = \(\frac{-1}{2}\) * (x - 2)
=> 2y - 6 = -x + 2
=> x + 2y = 8 is the equation of the line parallel to 2x + 4y = 5 and passes through the point (2,3)
=> Parallel line will have the same slope = \(\frac{-1}{2}\)
=> Equation of the line is given by y - \(y_1\) = m * (x - \(x_1\))
=> y - 3 = \(\frac{-1}{2}\) * (x - 2)
=> 2y - 6 = -x + 2
=> x + 2y = 8 is the equation of the line parallel to 2x + 4y = 5 and passes through the point (2,3)
Solved Problem : Find equation of a line perpendicular to a line and passing through a point
Q5. Find the equation of the line which is perpendicular to the line 2x + 4y = 5 and passes through the point (2,3).
Sol 5: Equation of line 2x + 4y = 5 = \(\frac{-2}{4}\) = \(\frac{-1}{2}\)
=> Slope of perpendicular line, m will be given by m * \(\frac{-1}{2}\) = -1
=> m = 2
=> Equation of the line is given by y - \(y_1\) = m * (x - \(x_1\))
=> y - 3 = \([fraction]2[/fraction]\) * (x - 2)
=> y - 3 = 2x - 4
=> y = 2x - 1 is the equation of the line perpendicular to 2x + 4y = 5 and passes through the point (2,3)
=> Slope of perpendicular line, m will be given by m * \(\frac{-1}{2}\) = -1
=> m = 2
=> Equation of the line is given by y - \(y_1\) = m * (x - \(x_1\))
=> y - 3 = \([fraction]2[/fraction]\) * (x - 2)
=> y - 3 = 2x - 4
=> y = 2x - 1 is the equation of the line perpendicular to 2x + 4y = 5 and passes through the point (2,3)
Watch the following video to learn the Basics of Co-ordinate Geometry
Watch the following video to learn about Equation of a Line
Hope it helps!
Good Luck!
