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

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    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.
    Thank you for your help in advance
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    (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.
    Thank you for your help in advance
    How did you get the perpendicular gradient?

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    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.
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    (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.
    Thank you for your help in advance
    Your gradient for the perpendicular line is wrong. If two lines with gradients m and M are perpendicular then mM = -1.
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    (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.
    Thank you for your help in advance
    please post your workings
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    "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.
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    I see my mistake. Solved the question now. Thank you very much everyone
 
 
 
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