help please ^__^

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1) find an equation of the line that passes through the point (5,-5) and is perpendicular to the line y= 2/3 x + 5  write your answer in the form ax + by + c = 0  where a, b and c are intergers.

2) find an equation of the line that pases through the point (-2,-3) and is perpendicular to the line y = -4/7 x + 5  write your answer in the form ax + by +c = 0  where a, b and c are intergers.

 could you  explain how u did it thnx XDD

Posted: 22-10-12 19:41 by Waqar Khanny :)

1) the gradient is 2/3 so the perpendicular gradient will be -3/2

to work out the equation of the line you need to use this formula: y - y₁  = m (x - x₁)(this is an equation of a straight line)

(y is the y-point,  is the gradient and x is the x-point)

so substitute the values in:  y - -5 = -3/2 (x - 5)

y + 5 = -3/2x + 15/2

then times both sides by 2 to get rid of the fractions and get integers:

2(y + 5) = -3x + 15

2y + 10 = -3x + 15

collect the terms: 2y = -3x + 5

to get it into the form ax + by + c = 0, move all the terms on to the left side

2y + 3x - 5 = 0

Posted: 22-10-12 21:04 by Joanne

thanx sooo much ^__^

Posted: 22-10-12 21:09 by Waqar Khanny :)

then do the same for the second one:

2) y - y = m (x - x)

(the perpendicular gradient of -4/7 is 7/4 - you take the reciprocal and swap its sign - to +)

y - -3 = 7/4 (x - -2)

y + 3 = 7/4x + 14/4

4 (y + 3) = 7x + 14

4y + 12 = 7x + 14

4y - 7x - 2 = 0

Posted: 22-10-12 21:10 by Joanne

Equation of a line y-y1=m(x-x1) is the same as y=mx+c but easier to solve as you can use one set of co-ordinates and substitute the values in y1 and x1 as long as you know the gradient, otherwise you work out the gradient using 2 sets of co-ordinates.

1) y=2/3x+5

Gradient equals=2/3 so the gradient for the line which is perpendicular to this and passes through (5,-5) equals -3/2. Remember, when you look for the gradient of a perpendicular line, you flip the values so 2/3 becomes 3/2 and change the sign so + becomes -.

As you have the gradient now, you can use the formula y-y1=m(x-x1) to find the equation. So, (5,-5), when substituted becomes y-(-5)=-3/2(x-5). Solve the equation: y+5=-3/2x -15/2 this equalts to y=-3/2x-25/2

As you have been asked to give the answer in the form of ax+by+c=0, you will have to times the equation by 2 to get rid of the fractions. This gives you 2y=-3x-25. Reorganise the equation: 2y+3x+25=0. 

Thsi si the method, do the same for question 2. BE sure to check the working out becuase I might not be correct, it's hard typing. But the method is right so you should be able to figure it out. Plus at the end to make sure both lines are perpendicular multiply the gradient of both lines to make sure you get -1. 

m1 x m2 = -1

m=gradient. 

Posted: 22-10-12 21:18 by Kelly smith

thanx bth of u seriously saved me ^__^ XDD

Posted: 22-10-12 21:20 by Waqar Khanny :)

No problem!

Posted: 22-10-12 21:23 by Joanne

^__^

Posted: 22-10-12 21:23 by Waqar Khanny :)