find the slope, mid point, equation, distance, parallel eqn, perpendicular eqn of...

[Aidan12321]

Ok so I am trying to find the slope, mid point, equation, distance, parallel equation, and perpendicular equation of points (-1,4) and (1,8)

 

[Deleted member 4993]

Ok so I am trying to find the slope, mid point, equation, distance, parallel equation, and perpendicular equation of points (-1,4) and (1,8)

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

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[stapel]

Ok so I am trying to find the slope, mid point, equation, distance, parallel equation, and perpendicular equation of points (-1,4) and (1,8)

Points have no slopes, etc. I'll guess that you mean "of the line segment with this points as endpoints".

Your class gave you a formula for the slope of the line through two points. (here) You plugged the given points into the given formula, simplified, and... then what?

Your class gave you a formula for the midpoint between two other points. (here) You plugged the given points into the given formula, simplified, and... then what?

Your class gave you a formula for the distance between two points. (here) You plugged the given points into the given formula, simplified, and... then what?

Your class gave you various formulas for the equations of straight lines through points. (here) Your class also gave you a relationship between the slopes of parallel lines. You picked one of these formulas, plugged either the points or else a point and the slope into it, simplified, and... then what?

Your class gave you a relationship between the slopes of parallel and perpendicular lines. How did you use this to find the perpendicular-line equation?

What does your book mean by "equation" for the two points (since clearly it doesn't mean "equation of the parallel line" or "equation of the perpendicular line")?

Please show all of your work. Thank you!

 

[hoosie]

Example: Finding equation of line passing through two given points

Hi Aidan - do you remember the general equation of a straight line y = mx + c ?
m = gradient (slope) of line c = y intercept
EXAMPLE:
If a straight line has gradient 3 and cuts the Y axis at -4 then equation of this line is:
y = 3x - 4
Your question doesn't give the slope of the line nor the y intercept so we have to use another form of the general equation of a straight line: y - y1 = m(x - x1) where m = gradient = rise/run and (x1, y1) are
co-ordinates of a point line passes through ( you can choose either point)
EXAMPLE:
If a straight line passes through points (2,-1) and (-2,7) then equation of line is given by:
y - y1 = m(x - x1) where (x1,y1) = (2,-1) and m = rise/run (height of step/width of step)
Therefore m = (y2 - y1)/(x2 - x1)
= (7 - -1)/(-2 - 2)
= 8/-4
= -2
We know x1 = 2 and y1 = -1
Substituting values of m = -2, x1 = 2 and y1 = -1 into general equation gives:
y - -1 = -2(x - 2)
y + 1 = -2x + 4
y = -2x + 3
To prove that we are correct we can substitute co-ordinates of second point (-2,7) into equation and show that LHS = RHS
7 = -2(-2) + 3
7 = 4 + 3
7 = 7
So y = -2x + 3 is the equation of the straight line

Note using y = mx + c this line intersects the Y axis at 3

Hope this example helps you to answer two parts of your question:
1) Gradient of straight line 2) Equation of straight line

Log in tomorrow for examples to help you work out the other parts of the question.








trying to find the slope, mid point, equation, distance, parallel equation, and perpendicular equation of points (-1,4) and (1,8)

 

[hoosie]

Finding equations of straight lines parallel and perpendicular to a given line

EXAMPLE:

Consider the equation of the straight line y = -2x + 3 where slope(m) = -2 and y-intercept(c) = 3
If we raise this line 2 units vertically upwards (parallel to Y axis) the equation becomes:
y = -2x + 5 (simply add 2 to y-intercept)

NB: When we multiply the gradients of two perpendicular lines the answer is -1
i.e m1Xm2 = -1 where m1 is slope of first line, m2 is slope of second one

For the line y = -2x + 3 m1 = -2 Since -2 x 1/2 = -1, m2 = 1/2
Now let's choose a point which the original line and perpendicular line both pass through.
For first line y = -2x + 3 when x = 1, y= -2(1) + 3 => y = 1 So line 1 passes through (1,1).
If line 2 also passes through point (1, 1) then its equation {using y - y1 = m(x - x1)} is given by:
y - 1 = (1/2)(x - 1) where (x1, y1) = (1,1) and m = 1/2 [note: m2 is the slope(m) in this case]
Therefore y = x/2 -1/2 +1 => y = x/2 + 1/2 is equation of perpendicular line.
We can check this is the correct answer by substituting co-ordinates (1,1) into equation.
If y = 1, LHS = 1 RHS = 1/2 + 1/2 => RHS = 1

 

[hoosie]

Finding midpoint of and distance between two given points

EXAMPLE:
Consider the two points (-3,2) and (5,7) where (x1,y1) = (-3,2) and (x2,y2) = (5,7)
To work out the co-ordinates of the midpoint we simply average the 2 x values and the 2 y values.

NB: Midpoint = [(1/2)(x1 + x2), (1/2)(y1 + y2)]

By substitution: Midpoint = [(1/2)(-3 + 5), (1/2)(2 + 7)]
Therefore midpoint = (1, 4.5)

NB: Distance between two points (x1, y1) and (x2, y2) is given by:
d = [(x2 - x1)^2 + (y2 - y1)^2]^(1/2)
By substitution d = [(5 - -3)^2 + (7 - 2)^2]^ (1/2)

[FONT=MathJax_Math-italic]d[FONT=MathJax_Main]= [/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main])^2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main])^2[/FONT][/FONT]

[FONT=MathJax_Math-italic]d[FONT=MathJax_Main]= 89[/FONT][/FONT]

[FONT=MathJax_Math-italic]d[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]9.434[/FONT][/FONT]