maths homework problem, i can do most of it, but then for some reason i get stuck?

Question

find the equation of the line that passes through the point of intersection of the lines x-3y=4 and 3x+y=2 and is perpendicular to the line 3x+4y=0. Give your final answer in the form ax+by+c=0 where a,b and c are integers to be determined.


its the last bit i'm lost on:

i got:

y=17

and

x=-5

so...the point of intersection is (-5,17)

and i know that the new line must be perpendicular to... so it will be -1/3x but then what??????

cheers

Answer 1

find the equation of the line that passes through the point of intersection of the lines x-3y=4 and 3x+y=2 and is perpendicular to the line 3x+4y=0. Give your final answer in the form ax+by+c=0 where a,b and c are integers to be determined.

first we find the point of intersection

x-3y=4,3x+y=2

>>>>x=1,y= -1

therefore, point of intersection of two lines
is(1,-1)

3x+4y=0
y=(-3/4)x

the gradient of line perpendicular to this line
= -1/gradient

perpendicular line passes through (1,-1) and
has a gradient of 4/3

equation of perpendicular is

y=4/3x+c say

substitute values at point(1,-1)

-1= 4/3+c

>>>>>c= -7/3

equation of line is y=4/3x-7/3

>>>>>3y-4x+7=0

therefore, the equation of the line that passes through the point of intersection of the lines x-3y=4 and 3x+y=2 and is perpendicular
to the line 3x+4y=0 is

3y-4x+7=0

when doing these problems,it helps to draw
a rough graph of the situation

another answer could be

4x-3y-7=0 or 300y-400x+700=0

i hope that this helps

Answer 2

A point of intersection means there is only one x-value and one y-value that fits both equations at the same time, so they must be treated as simultaneous equations.

(A)... x - 3y = 4
(B)... 3x + y = 2

Multiply (A) through by 3 to give (C).
(C)... 3x - 9y = 12
Subtract (C) from (B) to give :
10y = -10
so, y = -1

With y = -1 in (A), for example, x - 3(-1) = 4,
so, x = 1
Thus, the point of intersection is (x, y) = (1, -1).

The equation to your line is y = mx + c.
We know that (1, -1) is on the line, so :
-1 = m(1) + c, or
(D)... c = -m -1, where c is the y-intercept and m the slope.

Now the line, 3x + 4y = 0 is equivalent to y = (-3/4)x + 0.
The slope of this line is therefore -3/4.

For your line to be perpendicular to this one, the slope of your line must be the negative reciprocal of -3/4, which is +4/3.
Thus, in (D), m = 4/3, therefore, c = -4/3 - 1 = -7/3.

The equation to your line is then : y = (4/3)x - 7/3.
Rearranging this gives : 4x - 3y - 7 = 0.

Answer 3

Why not just draw it on a piece of paper? If you do you will see that misty is quite right that the two lines cross at (x=1,y=-1) and dsee that the line 3x+4y=0 drops 4 units in y for every three units increase in x. Draw a line at right angles through this and see that any line of the form 4x-3y=constant is a solution. Now find the constant that lets this line pass through x=1,y=-1

4(1)-3(-1)=constant = 7

4x-3y+7=0 Rats! I got it wrong right at the end!

4x-3y-7=0 what an idiot!

Answer 4

You are a bit wrong...
Put equations in "normal" format (y=ax+b)
1. x -3y = 4 -> y = 1/3x - 4/3 -> m1=1/3, b1=-4/3
2. 3x + y = 2 -> y = -3x +2 -> m2=-3, b2=+2
Find the intersection (xi, yi):
xi = (b2-b1) / (m1-m2) = 1 (you can do the arithmetics!)
yi = m1.xi + b1 = -1
Analyse 3x +4y = 0
or y = -3/4 x + 0
It is a line that passes through origin (b=0)
with an inclination m = -3/4
The inclination of the normale to that line is
mn = -1 / m = +4/3
Now you can find the equation of a line that passes through a point (xi, yi) with an inclination of mn:
y - yi = m(x - xi) -> y = 4/3x - 7/3
Finally, we put it in the format requested (ax+by+c=0)
3y = 4x -7
-4x+3y+7 = 0
==========
Solution verified graphically.
Thanks for the problem: I'll use it in my next class!
You can find the inclination "m" of that line:

Answer 5

u have it wrong .. first solve the first 2 eqn.s too get the point thru which the lines pass iu get x=1 y= -1 , then u gettthe slope of the line perpendicular to the line which will -3/4 . now you u can getthe slope of the line who eqn. u need it will 4/3
now use the equation y=mx+c to getthe value of c subatitutinf the value of the slope in place of m and x=1, y= -1 as calcuated earlier . you get c= -7/3. so you get your final equation

(1)x + (-1) y + (-7/3) =0
on solving you get ,

3x -3y-7=0

thats your answer ..

Hope you understood .

Answer 6

ok...so you got the point right [i guess because i didnt check it myself].
when a line is perpendicular to other line, its gradient is a negative reciprocal of the 'other line' in this case 3x+4y=0
you solve this for y=-3/4x
therefore the gradient of the line you are looking for is 4/3
now use the formula:
y-y1=m[x-x1]
y-17=4/3[x+5]
y-17=4/3x + 20/3
y-17-4/3x-20/3=0
3y-51-4x-20=0
3y-4x-71=0
4x-3y+71=0

thats it

Answer 7

y-y1= m(x-x1)
y-17 = -1/3(x+5)
y-17 = -x/3 - 5/3
y = -x/3 - 5/3 + 17
y = -x/3 +46/3

Thats using ur gradient but it should be
3x +4y = 0
y = -3x/4
.'. m = 4/3