Coordinates,Lines & Points 1....?

Question

P is the point (7,5) & a1 is the line with the equation 3x+4y=16

1) Find the equation of the line a2 which passes through P & is perpendicular to a1.
2) Find the point of intersection of the lines a1 & a2.
3) Find the perpendicular distance of P from the line a1.

i'm really stuck on these,can anyone explain how to get the answers to this pls with the theories & equations stated clearly?thanks...

Answer 1

Implicit form: 3x+4y=16

You need the explicit form:

4y=-3x+16

Explicit form: y=-(3/4)x+4, this is a1.

Explicit form of a2: y=mx+q

When two lines are perpendicular, their m coefficients product = -1, that is:

m*m'=-1

So m of a2 is 4/3.

y=(4/3)x+q.

Now you need the q value: just put in the equation the point values:

5=(4/3)*7+q

15=28+3q

3q=15-28

3q=-13

q=-(13/3)

1) a2: y=(4/3)x-(13/3)
3y=4x-13
4x-3y=13.

2) You have to put the equation in a system.
You can multiply a1 for 4 and a2 for (-3). Then you can sum the equations side by side:

a1: 12x+16y=64
a2: -12x+9y=-39

12x-12x+16y+9y=64-39

25y=25

y=1.

Choose one equation:

3x+4*1=16

3x=16-4

3x=12

x=4.

2) the point is (4,1).

3) Be x0=7, y0=5, a=3, b=4, c=-16

The formula is:

d= |a*x0 + b*y0 + c|/sqrt(a^2 + b^2)

The numerator is:

|3*7+4*5-16|=|21+20-16|=|41-16|=25

The denominator is

sqrt(9+16)=the square root of 25, that is 5.

3)The distance is d=5.

Answer 2

Coordinates,Lines & Points 1....?
P is the point (7,5) & a1 is the line with the equation 3x+4y=16

1) Find the equation of the line a2 which passes through P & is perpendicular to a1.
Ans:
A line perpendicular to the given line is obtained by interchanging the cofficents of x and y and also sign in between them thus a line perpendicular to a1 or line a2 will be
4x -3y =k......................i
Since point P(7,5) lies on i, therefore
4(7) -3(5)=k
28 - 15 = k
13 = k....................ii
Therefore the equation of line a2 is
4x - 3y = 13........................iii
2) Find the point of intersection of the lines a1 & a2.
Ans:
By solving the i and iii we get
3x +4y=16.....................i ]multiply by 3
4x -3y = 13...................iii ] multiply by 4
9x + 12 y = 48..............iv
16x - 12y = 52..............v
adding iv and v
25x =100 or x = 4 ........vi
from i and vi we get y=1...................vii
Therefore the point of intersection is (4,1)
3) Find the perpendicular distance of P from the line a1.
Ans:
Perpendicular distance from point(7,5) to the given line will be
obtained by changing x with 7 and y with 5 and shift 16 on the left side and whole is divided by sqrt of (3^2+4^2)= 5
p=perpendicular distance= {3(7) +4(5) - 16}/ 5
p=25/5=5 units
Hence perpendicular distance from point P(7,5) on line a1 is 5 units

Answer 3

let the point on the eq be (x,y)
then
as they r perpendicular then the product of there slopes will be -1
slope of a2 is =4/3 (y2-y1/x2-x1)
slope of a1 is -3/4 (4y=-3x+16)then(y=-3/4x+16)
y-5=4/3(x-7)
y-5/x-7=4/3
cross multiply
3y-15=4x-28
4x-3y=13(ans for 1)




2.point of intersection is
solve eq
4x-3y=13 &
3x+4y=16
we get
x=4 & y=1
point is (4,1)(ans for 2)



3.perpendicular distance is
taking eq 3x+4y-16=0
modulas(3(7)+4(5)-16/square root(3^2+4^2))
after solving
distance is 5(ans for 3)


:-)

Answer 4

1.slope of the given equation=-3/4
slope of theperpendicular line=4/3
passes through (7,5)
equation=y-5=(4/3)(x-7)
3y-15=4x-28
4x-3y-13=0
4x-3y=13
3x+4y=16

2.12x-9y=39
12x+16y=64
25y=25 so y=1
substitutingx=4
so pointof intersection is (4,1)

3.distance=Ax1+by1+C/(A^2+B^2)^1/2
3(7)+4(5)-16/(3^2+4^2)^1/2
21+20-16/5=5 units

Answer 5

This really is some fairly basic coordinate geometry. If you're having a bad time with this, then you need to get a tutor *right* *now* to start working with you because it's only gonna get deeper from here on in.


Doug

Answer 6

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