four questions that i have no idea how to do :
find the equation of the straight line that:
- passes through points (-2,8) and (4,2)
-is parallel to the line 2y-6x+12=0
-is perpendicular to the line y=-2x+15
and
the equation of the line joining A and D [ A(2,5) , D(0,7) ]
if you could show how you worked it out that would be great to because this is in my test tomorrow :/ thanks <3
find the equation of the straight line that:
- passes through points (-2,8) and (4,2)
-is parallel to the line 2y-6x+12=0
-is perpendicular to the line y=-2x+15
and
the equation of the line joining A and D [ A(2,5) , D(0,7) ]
if you could show how you worked it out that would be great to because this is in my test tomorrow :/ thanks <3
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The points (-2,8) and (4,2) gives you the information that when
Series 1 ) x=-2 y=8
and
Series 2) when x=4 y =2
Use the equation y-y1=m(x-x1) and find an equation for both these points
Series 1)
y-8=m(x+2)
y=m(x+2)+8
Series 2)
y-2=m(x-4)
y=m(x-4)+2
Set these to equal one another and solve for m.
m(x+2)+8=m(x-4)+2
mx+2m-mx+4m+8-2=0
2m+4m+6=0
6m=-6
m=-1
Plug this into either equation found above and simplify
y=m(x+2)+8
y=-1x-2+8
y=-1x+6
f they're parallel it means that their slopes are equal, and if they're perpendicular it means their slopes multiplied by one another equals -1.
So put line 2y-6x+12=0 into y=mx+b form
2y=6x-12
y=3x-6
Slope = 3
So you know that the slope of the line that is parallel to this has a slope of 3.
So you get y=3x+b, where b can be anything except -6
Perpendicular
y=-2x+15
you have the slope y=-2, and the line you're looking for has the slope x, where x*-2=-1.
Solving this you get x=1/2
finally you get the equation
y=.5x+b where b is anything
Same as the first one, the line passes through point (2,5) and (0,7)
y-y1=m(x-x1) for both of them and solve for m then simplify.
Series 1 ) x=-2 y=8
and
Series 2) when x=4 y =2
Use the equation y-y1=m(x-x1) and find an equation for both these points
Series 1)
y-8=m(x+2)
y=m(x+2)+8
Series 2)
y-2=m(x-4)
y=m(x-4)+2
Set these to equal one another and solve for m.
m(x+2)+8=m(x-4)+2
mx+2m-mx+4m+8-2=0
2m+4m+6=0
6m=-6
m=-1
Plug this into either equation found above and simplify
y=m(x+2)+8
y=-1x-2+8
y=-1x+6
f they're parallel it means that their slopes are equal, and if they're perpendicular it means their slopes multiplied by one another equals -1.
So put line 2y-6x+12=0 into y=mx+b form
2y=6x-12
y=3x-6
Slope = 3
So you know that the slope of the line that is parallel to this has a slope of 3.
So you get y=3x+b, where b can be anything except -6
Perpendicular
y=-2x+15
you have the slope y=-2, and the line you're looking for has the slope x, where x*-2=-1.
Solving this you get x=1/2
finally you get the equation
y=.5x+b where b is anything
Same as the first one, the line passes through point (2,5) and (0,7)
y-y1=m(x-x1) for both of them and solve for m then simplify.
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You will be using these two formulas for problems 1 and 4:
· Point – slope formula: [where the two points are (x1, y1) and (x2, y2)
y - y1 = m (x - x1)
·
· 2 point given formula:
y - y1 = m (x - x1) and m = (y2 – y1) / (x2 – x1)
Just substitute the coordinates of the two points into the equation. Your final answer should probably be in the form of: y = mx + b
For #2 and #3, you need to get the given equations in the form of: y = mx + b.
Then for #2, remember that the slope of a line that is parallel to another has the same slope.
For a perpendicular line in #3, you need a slope that is the negative reciprocal. ( 2/3 ---> -3/2 )
· Point – slope formula: [where the two points are (x1, y1) and (x2, y2)
y - y1 = m (x - x1)
·
· 2 point given formula:
y - y1 = m (x - x1) and m = (y2 – y1) / (x2 – x1)
Just substitute the coordinates of the two points into the equation. Your final answer should probably be in the form of: y = mx + b
For #2 and #3, you need to get the given equations in the form of: y = mx + b.
Then for #2, remember that the slope of a line that is parallel to another has the same slope.
For a perpendicular line in #3, you need a slope that is the negative reciprocal. ( 2/3 ---> -3/2 )