# Thread: Coordinate Geometry

1. ## Coordinate Geometry

I had this big assignment to do, and I did most of the questions except for these few (or I did them and I'm not sure if that's how I go about working them out)

1. State the y_intercept of the line: 3x+4y = -3 with equations.

I did: 4y = 3x+8 / y_int = -8?

2. Show that (2, 1) lies on the same line with the equation y=2x+3

I started with: 2 x 2 + -3x1 = -3

3. Parallel to 3x-4y=2 passing through (1, 2)

I did: 3x-4y =3 x 1 - 4 x 1
=3x-4y=-1

4. Perpendicular to 3x-4y = 2 passing through (1, 2)

Absolutely no idea how to do this.

5. Two lines have slopes k+2=-1. Find k given that:
a. The lines are parallel
b. The lines are perpendicular

I got:

a. k+2 = -1
b. -1/2k

Can anyone please help me out with these?

2. Originally Posted by suckatmaths
...

1. State the y_intercept of the line: 3x+4y = -3 with equations.

I did: 4y = 3x+8 / y_int = -8?
Where does the +8 com from?
You have to plug in x = 0 to calculate the y-intercept.

2. Show that (2, 1) lies on the same line with the equation y=2x+3

I started with: 2 x 2 + -3x1 = -3
Where did you find this -3 ? Plug in the coordinates of the point and check if you get a true statement. In your case:
$\displaystyle 1 = 2 \cdot 2 +3~\implies~1 \neq 7$ that means the point doesn't belong to the line.
3. Parallel to 3x-4y=2 passing through (1, 2)

I did: 3x-4y =3 x 1 - 4 x 1
=3x-4y=-1
Why do you use the x-value in the place of the y-variable? Your method is OK but you should have gotten:
$\displaystyle 3 \cdot 1 - 4 \cdot 2 = -5$ and therefore the equation of the parallel line is $\displaystyle 3x - 4y = -5$
4. Perpendicular to 3x-4y = 2 passing through (1, 2)
If a straight line has the equation
$\displaystyle ax +by = c$ then the perpendicular line has the equation
$\displaystyle bx - ay = d$
So you know now the LHS of the equation. Find the value of d using the same way as in #3
Absolutely no idea how to do this.

5. Two lines have slopes k+2=-1. Find k given that:
a. The lines are parallel
b. The lines are perpendicular

I got:

a. k+2 = -1
b. -1/2k

...
to a): Two lines are parallel if the slopes $\displaystyle m_1$ and $\displaystyle m_2$ are equal. Thus $\displaystyle m_1=m_2~\implies~k+2=-1~\implies~k=-3$

to b) Two lines are perpendicular to each other if the slopes satisfy the equation
$\displaystyle m_1 \cdot m_2 = -1$

I'll leave the rest for you.

3. 1. State the y_intercept of the line: 3x+4y = -3 with equations.
I did: 4y = 3x+8 / y_int = -8?
rewrite the equation with y as subject and its coefficient always = +1 ,
hence y = mx + c, where c is the y-intercept, and m is the slope of line.
3x + 4y = -3
4y = -3x -3
y = $\displaystyle -\frac{3}{4}x - \frac{3}{4}$
so y-intercept = $\displaystyle -\frac{3}{4}$

2. Show that (2, 1) lies on the same line with the equation y=2x+3
I started with: 2 x 2 + -3x1 = -3
substitute (2,1) into y = 2x + 3 :
left hand side = 1
right hand side = 2(2) + 3 = 7
since left hand side is not equal to right hand side, the point (2,1) does not lie on y = 3x + 3 ........check your question again

3. Parallel to 3x-4y=2 passing through (1, 2)
I did: 3x-4y =3 x 1 - 4 x 1
=3x-4y=-1
What you presented is wrong. (1,2) lies on a line that is parallel to another line 3x - 4y = 2. it does not mean (1,2) simply lies on this 3x - 4y = 2.
Being parallel means the two lines share the same slope, m.
3x - 4y = 2
3x - 2 = 4y
y = $\displaystyle \frac{3}{4}x - \frac{1}{2}$
hence its slope = $\displaystyle \frac{3}{4}$
Therefore, the equation of the line parallel to 3x - 4y = 2 and passes through (1,2) is:
$\displaystyle \frac{y - 2}{x - 1} = \frac{3}{4}$
In the end you get 7x - 4y + 1 = 0

4. Perpendicular to 3x-4y = 2 passing through (1, 2)
Absolutely no idea how to do this.
Similar to question 3 but here the slope of the line you are working on is -\frac{4}{3} because the product of the slopes of two lines, perpendicular to each other, is equal to -1.

5. Two lines have slopes k+2=-1. Find k given that:
a. The lines are parallel
b. The lines are perpendicular
I got:
a. k+2 = -1
b. -1/2k
Don't understand what does k = -3 stand for?...

4. Originally Posted by ukorov
...
3. Parallel to 3x-4y=2 passing through (1, 2)
I did: 3x-4y =3 x 1 - 4 x 1
=3x-4y=-1
What you presented is wrong. (1,2) lies on a line that is parallel to another line 3x - 4y = 2. it does not mean (1,2) simply lies on this 3x - 4y = 2.
Being parallel means the two lines share the same slope, m.
3x - 4y = 2
3x - 2 = 4y
y = $\displaystyle \frac{3}{4}x - \frac{1}{2}$
hence its slope = $\displaystyle \frac{3}{4}$
Therefore, the equation of the line parallel to 3x - 4y = 2 and passes through (1,2) is:
$\displaystyle \frac{y - 2}{x - 1} = \frac{3}{4}$
In the end you get 7x - 4y + 1 = 0

...
I don't want to pick at you but you must have made a small mistake:

$\displaystyle \frac{y - 2}{x - 1} = \frac{3}{4}~\implies~4(y-2)=3(x-1) ~\implies~ 4y-8=3x-3~\implies~-3x+4y-5=0$

5. Originally Posted by earboth
I don't want to pick at you but you must have made a small mistake:

$\displaystyle \frac{y - 2}{x - 1} = \frac{3}{4}~\implies~4(y-2)=3(x-1) ~\implies~ 4y-8=3x-3~\implies~-3x+4y-5=0$
...foolish mistake

6. Originally Posted by ukorov
...foolish mistake
It is customary among the regulars to use the thanks button when someone spots an error in one of your posts.

CB