Find the ratio in which the staight line 5x+4y=4 divides the join of (4,5) and (7,-1)?

2 Answers
Feb 17, 2018

1/3

Explanation:

There are a few things we'll need to establish before we can find the requested ratio. If we want a ratio between two line segments, we'll obviously need two line segments, and in order to have two line segments, we'll need three points. Two points - A(4,5) and B(7,-1) - have been provided, but we'll need to find the third by determining where the line 5x+4x=4 and AB intersect.

Before we can do that, though, we'll need to find the equation for AB. If you'll recall, a line is defined by some constant rate of change - its slope, m, were

m=(Deltay)/(Deltax)

Here, we can take our two points to find that

m=(-1-5)/(7-4)=-6/3=-2

With that in place, we can derive the general form of the line's equation by fixing one point in place - say, A(4,5) - and letting the other vary (i.e. it can be any arbitrary point (x,y) on the line):

-2=(y-5)/(x-4)

-2(x-4)=y-5
-2x+8=y-5
13=y+2x
2x+y=13

Now that we have that equation, we can set up a system and solve for the intersection of the two:

(1)\ 5x+4y=4
(2) \ 2x+y=13

Here, I'll solve by substition by first manipulating the second equation to get:

y=13-2x

which, substituting into (1):

5x+4(13-2x)=4
5x+52-8x=4
52-3x=4
48=3x
x=16

Solving for y:

y=13-2(16)
y=13-32
y=-19

So our point of intersection is C(16,-19).

For a feel as to what our picture looks like at this point:

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We're almost done. Our final steps are to find the length of AB, find the length of BC, and finally, to find (AB)/(BC).

We can use the distance formula to calculate both of those distances, obtaining

AB=sqrt((7-4)^2+(-1-5)^2)=sqrt(3^2+(-6)^2)=sqrt(9+36)
=sqrt9sqrt(1+4)=3sqrt5

BC=sqrt((16-7)^2+(-19-(-1))^2)=sqrt(9^2+(-18)^2)
=sqrt(9^2+9^2(-2)^2)=sqrt(9^2)sqrt(1+(-2)^2)=9sqrt(5)

And a ratio between the two of:

(AB)/(BC)=(3sqrt(5))/(9sqrt(5))=3/9=1/3

Feb 17, 2018

Hence, the ratio in which the point divides the line is
4:-3=-4:3

Explanation:

We need to find the point of intersection of the lines
5x+4y=4 and
(y-5)/(x-4)=(-1-y)/(7-x)

Simplifying the second line

(y-5)(7-x)=(-1-y)(x-4)

(y-5)(x-7)-(y+1)(x-4)=0

Expanding
xy-7y-5x+35-(xy-4y+x-4)=0

xy-7y-5x+35-xy+4y-x+4=0

Rearanging and simplifying
-6x-3y+39=0
Dividing by 13
-2x-y+13=0

or

2x+y=13

The equations are:

5x+4y=4-------(1)
2x+y=13-------(2)

detA=5xx1-2xx4=5-8=-3
detx=4xx1-13xx4=4-52=-48
dety=5xx13-2xx4=65-8=57

x=detx/detA=-48/-3=16
y=dety/detA=57/-3=-19

(x,y)-=(16,-19)

Check

5x+4y=5xx16+4xx(-19)=80-76=4

2x+y=2xx16+(-19)=32-19=23

Verified.

Hence, the intersection point is P-=(16,-19)
One end of the line is
Other end of the line is B-=(7,-1)
Arranging in the form of APB
(x_A,y_A)-=(4,5)
(x_P,y_P)-=(16,-19)
(x_B,y_B)-=(7,-1)
P divides AB in the ratio
(x_P-x_A)/(x_B-x_P)=(16-4)/(7-16)=12/-9=4/-3along x axis
(y_P-y_A)/(y_B-y_P)=(-19-5)/(-1-(-19))=(-24)/18=(-4)/3 along y axis
Check:
Both are same
Justifying the coordinate for intersection point

Hence, the ratio in which the point divides the line is
4:-3=-4:3