Given the line 2x-3y=9 and the point (4,-1), find lines through the point...

[zaheer_abbas]

Given the line 2x-3y=9 and the point (4,-1), find lines through the point that are (a) parallel to the given line and (b) perpendicular to it.

 

[zaheer_abbas]

Show your work. You posted another similar problem.

i have no idea to solve this problem i try many time i'm week in math but unfortunately i'm study,

 

[HallsofIvy]

You titled this "I need to solve this problem" and you are right- It would do you no good for someone else to do it for you. Here are the things you need to think about:

What is the slope of the line given by 2x- 3y= 9?

What is the slope of a line parallel to that line?

What is the slope of a line perpendicular to that line?

How do you find the equation of a line if you know its slope and one point on the line?

Whoever gave you this problem clearly expects you to know those things.

 

[stapel]

i have no idea to solve this problem

So you've never been taught this material? Why then are you being assigned questions on this material?

i try many time

Great! Please reply showing at least two of your attempts, so we can see where you're having problems, and provide lesson links so you can begin to learn this material. Please be complete. Thank you!

 

[hoosie]

Worked example for finding equation of straight line given its gradient and a point

Given the line 2x-3y=9 and the point (4,-1), find lines through the point that are (a) parallel to the given line and (b) perpendicular to it.

Hi Zaheer - here is an example that should help you understand how to answer your question.


We will find the equation of straight line which passes through (2,5) and is parallel
to line y = 2x + 3.
Given line y = 2x + 3 is of form y = mx + c where m = gradient, c = y - intercept
⇒ m = 2, C = 3
Straight line with known gradient and point through which it passes can be written in form:
y - y1 = m(x - x1) where m = gradient and (x1, y1) = point
We know m =2, x1 = 2, y1 = 5
Substituting in these values gives: y - 5 = 2(x - 2) ⇒ y - 5 = 2x - 4 ⇒ y = 2x + 1
To find the equation of a straight line which passes through (2,5) but is perpendicular to
line y = 2x + 3 you need to remember that the gradients of perpendicular lines multiply
to equal -1.
For the required line we have
m= -1/2, x1 = 2, y1 = 5
Substituting in these values gives: y - 5 = (-1/2)(x - 2) ⇒ y - 5 = -x/2 + 1 ⇒ y = -x/2 + 6
To find the equation of a straight line which passes through (2,5) and is parallel to
line 3x + 4y = 12 we need to rearrange the equation of the given line to make y the subject
(y must be on its own). Rearranging gives: 4y = -3x + 12 (subtracting 3x from both sides)
⇒ y = -3x/4 + 3 (dividing both sides by 4)
For the required line we have
m= -3/4, x1 = 2, y1 = 5 Substituting in these values gives: y - 5 = (-3/4)(x - 2)
⇒ y - 5 = -3x/4 + 3/2 ⇒ y = -3x/4 + 13/2
To find the equation of a straight line which passes through (2,5) but is perpendicular
to line 3x + 4y = 12
we substitute in the values m= 4/3 (since -3/4 X 4/3 = -1) , x1 = 2, y1 = 5 to give:
y - 5 = (4/3)(x - 2) ⇒ y - 5 = 4x/3 - 8/3 ⇒ y = 4x/3 + 7/3

Important points to understand:
* Gradient of straight line (m) = rise/run
e.g if m = 2 = 2/1 this means as x goes up by 1, y goes up by 2
(line goes up from left to right for a positive gradient)
e.g if m = -2/3 this means as x goes up by 3, y goes down by 2
(line goes down from left to right for a negative gradient).


* When the equation of a straight line is written in the form y = mx + c e.g y = 2x + 3 (where y is on its own), the gradient (m) is the number in front of x (in this case 2) and the y-intercept is the last number (in this case 3).


* When the equation of a straight line is written in a different way e.g 2x - 3y = 6 (where y is not on its own), we must rearrange the equation to get y by itself so that we can work out the gradient.
We rearrange 2x - 3y = 6 to give 2x = 6 + 3y (adding 3y to both sides of equation)
It follows 2x - 6 = 3y (subtracting 6 from both sides) therefore 2x/3 -2 = y (dividing both sides by 3)
The gradient (m) = 2/3 and the y-intercept is -2


* The gradients of two lines that run parallel to each other are equal e.g y = -3x + 1 and y -3x + 6 are the equations of two parallel lines since both have a gradient of -3. If the first line is shifted up 5 (difference between y-intercepts) it will coincide with the second line.


* The gradients of two lines that run at right angles to each other multiply together to equal -1.
e.g y = 2x/3 - 5 and y = -3x/2 + 4 are equations of two lines which are perpendicular. First line has gradient (m1) = 2/3 and second line has gradient (m2) = -3/2. m1 X m2 = 2/3 X -3/2 = -1


* If we know the gradient of a straight line and the co-ordinates of a point it passes through,
we can use the following general equation to work out the equation of the line:
y - y1 = m(x - x1) where m = gradient and (x1, y1) = point
e.g straight line has gradient (m) = -3/4 and passes through (8, -1)
Substituting m = -3/4, x1 = 8, y1 = -1 into general equation gives:
y + 1 = -3/4(x - 8) ⇒ y + 1 = -3x/4 + 6 ⇒ y = -3x/4 + 5

 

[hoosie]

Hoosie, much easier to refer OP to sites that explain; like:
http://www.purplemath.com/modules/strtlneq3.htm

Like, why reinvent the wheel?

Prefer to use my teaching experience to assist students myself with good explanations.