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I need help understanding how to write an equation
write an equation of the given point and parallel to the given line.(6,-7) 7x - 6y = 5
I do not have the slightest clue where to even begin. I don't need it solved but just want to understand. Maybe someone could solve one similar to this one and show me how. I would be very greatful for any help. Thank you.
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- By jerrydoug06 |
- Category: algebra |
- Grade level: college and above
- May/18/2009 |
- Answers to this question: 2
|
Answer by: dawid66
When an equation is in Standard Form, Ax + By = C, the slope is -A/B
Given 7x -6y =5 has a slope of -7/(-6) which simplifies to 7/6
A line parallel to the given line has the same slope as the given line.
The new equation will have the same slope, m = 7/6
Use Point Slope Form, y-y1 = m (x-x1) when the slope, m, and a point is known.
y - (-7) = (7/6) (x -6)
This Point Slope Form can easily be changed over to Slope Intercept Form.
Simplify the equation to get y by itself.
y + 7 = (7/6)x - (7/6)6
y = (7/6)x -7 -7
y = (7/6)x -14
This is the equation of a line parallel to the original equation that goes through the given point, (6,-7)
I hope this helps and keep asking good questions like this.
MindZinger is a great site to get help and it is free.
Given 7x -6y =5 has a slope of -7/(-6) which simplifies to 7/6
A line parallel to the given line has the same slope as the given line.
The new equation will have the same slope, m = 7/6
Use Point Slope Form, y-y1 = m (x-x1) when the slope, m, and a point is known.
y - (-7) = (7/6) (x -6)
This Point Slope Form can easily be changed over to Slope Intercept Form.
Simplify the equation to get y by itself.
y + 7 = (7/6)x - (7/6)6
y = (7/6)x -7 -7
y = (7/6)x -14
This is the equation of a line parallel to the original equation that goes through the given point, (6,-7)
I hope this helps and keep asking good questions like this.
MindZinger is a great site to get help and it is free.
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- May/18/2009 |
- Answers by dawid66: 429 |
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Answer by: neela
First Answer
A line Parallel to ax+by+c =0 is of the form
ax+by+k = 0 (1) . Since this passes through the given point (x1, y\1) ,
a(x1)+b(y1) +k = 0 (2)
(1)-(2) gives: a(x-x1)+b(y-y1) =0 (3)
Replace in (3) , (x1 , y1) by (6, -7) , a=7, b= -6 and simplify to get a standard form:
7(x-6) +(-6)(y -(-7)) = 0
7x-42-6y -42 =0
7x-6y-84=0
ax+by+k = 0 (1) . Since this passes through the given point (x1, y\1) ,
a(x1)+b(y1) +k = 0 (2)
(1)-(2) gives: a(x-x1)+b(y-y1) =0 (3)
Replace in (3) , (x1 , y1) by (6, -7) , a=7, b= -6 and simplify to get a standard form:
7(x-6) +(-6)(y -(-7)) = 0
7x-42-6y -42 =0
7x-6y-84=0
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- May/18/2009 |
- Answers by neela: 222 |
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