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# C1 Finding equation of a straight line from point to the line watch

1. The question is: "The point A has coordinates (1,7) and the point B has coordinates (3, 1). The mid-point of AB is P. Find the equation of the straight line which passes through P and which is perpendicular to the line 5y + x = 7. Give you answer in the form y = mx + c."

I worked out P which is (2,4). Then I got the perpendicular gradient of 5y + x = 7 which would be -1/2. Then I used the point-slope form of the equation of the line which is y - y1 = m(x - x1). That got me the completely wrong answer.
The correct answer is y = 5x - 6.
2. (Original post by olegasr)
The question is: "The point A has coordinates (1,7) and the point B has coordinates (3, 1). The mid-point of AB is P. Find the equation of the straight line which passes through P and which is perpendicular to the line 5y + x = 7. Give you answer in the form y = mx + c."

I worked out P which is (2,4). Then I got the perpendicular gradient of 5y + x = 7 which would be -1/2. Then I used the point-slope form of the equation of the line which is y - y1 = m(x - x1). That got me the completely wrong answer.
The correct answer is y = 5x - 6.
How did you get the perpendicular gradient?

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3. First rearrange 5y + x = 7 to find the gradient then use the rule that lines that are perpendicular, their gradients multiply to give you -1.
Then you will have y=(new gradient,m)x + c. Put in the values that you got, 2 and 4 into y=mx+c. Find c and you should be good.
4. (Original post by olegasr)
The question is: "The point A has coordinates (1,7) and the point B has coordinates (3, 1). The mid-point of AB is P. Find the equation of the straight line which passes through P and which is perpendicular to the line 5y + x = 7. Give you answer in the form y = mx + c."

I worked out P which is (2,4). Then I got the perpendicular gradient of 5y + x = 7 which would be -1/2. Then I used the point-slope form of the equation of the line which is y - y1 = m(x - x1). That got me the completely wrong answer.
The correct answer is y = 5x - 6.
Your gradient for the perpendicular line is wrong. If two lines with gradients m and M are perpendicular then mM = -1.
5. (Original post by olegasr)
The question is: "The point A has coordinates (1,7) and the point B has coordinates (3, 1). The mid-point of AB is P. Find the equation of the straight line which passes through P and which is perpendicular to the line 5y + x = 7. Give you answer in the form y = mx + c."

I worked out P which is (2,4). Then I got the perpendicular gradient of 5y + x = 7 which would be -1/2. Then I used the point-slope form of the equation of the line which is y - y1 = m(x - x1). That got me the completely wrong answer.
The correct answer is y = 5x - 6.
6. "5y + x = 7 which would be -1/2"

Not quite. First rearrange into the form y = mx + c

Then find the gradient from there. Make sure you have only y on the left hand side and not 5y.
7. I see my mistake. Solved the question now. Thank you very much everyone

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