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Coordinate geometry...Find the ratio in which the line joining A(6,5) and B(4,-3) are...

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ramesh-chandr... | Student, Grade 10 | eNotes Newbie

Posted February 15, 2010 at 12:33 AM via web

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Coordinate geometry...

Find the ratio in which the line joining A(6,5) and B(4,-3) are divided by the line y=2. AB also intersects the x axis at P. Also find the coordinates of point P.

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giorgiana1976 | College Teacher | Valedictorian

Posted February 15, 2010 at 1:22 AM (Answer #1)

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First of all, we have to find out the equation of the line AB, in order to express the condition of intersecting 2 lines, which is: the coordinates of the point M, resulted from the intersection between lines AB and y=2, have to verify the equations of the 2 lines :AB and Y=2.

In order to find out the equation of the line AB:

(xB-xA)/(x-xA) = (yB-yA)/(y-yA)

Now, we'll substitute the values of known coordinates:

(4-6)/(x-6) = (-3-5)/(y-5)

-2/(x-6) = -8/(y-5)

After dividing by (-2):

1/(x-6) = 4/(y-5)

Cross multiplying:



y=4x-19, this being the eq. for the line AB.

The M point belongs to AB, only if it's coordinates verifies the equation of the line AB.


Also yM=2.

By substituting yM=2 into yM=4xM-19, the result will be:




M point coordinates are xM=21/4 and yM=2.

Now we have to find out the length of the segment AB and segment AM, in order to decide the ratio the line AB is split by the line y=2.

The AM segment's length:

AM= sqrt[(xM-xA)^2+(yM-yA)^2]

AM= sqrt[(21/4 - 6)^2 + (2-5)^2]

AM= sqrt[(9/16)+9]


AB= sqrt[(xB-xA)^2 + (yB-yA)^2]

AB= sqrt[(4-6)^2 + (-3-5)^2]

AB= sqrt(68)

AB= 2sqrt17

MB=AB-AM=2sqrt17 - 3sqrt17/4


The ratio is : AM/MB=(3sqrt17/4)/(5sqrt17/4)


To find out the coordinates for P, we have to remember again the condition that P coordinates has to verify the equation for AB line also.


But yP=0, because P is placed on x axis, so:




P coordinates are: xP=19/4 and  yP=0

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neela | High School Teacher | Valedictorian

Posted February 15, 2010 at 3:01 AM (Answer #2)

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The line Joining (a,b) and (c,d)) is given by:

y-c = (d-b)/(c-a)[x-a]

So, the line Joining A(6,5) and B(4,-3) is given by:

y-5 =(-3-5)/(4-6)(x-6) . Or

y = 4(x-6)+5..................(1).The line y = 2 cuts this line at x1 given by 2= 4(x1-6)+5. Or x1 = (2-5+24)/4 = 5.25. Therefore the coordinates of the points of intersection of the line y =2 and the line AB is at C(5.25,2).

Therefore, if the Ratio of AC and CB is k:1, then ki given by:

(k*4+1*6)/(k+1) = 5.25 or

4k+6=5.25k+5.25 Or

6-5.25= 5.25k-4k  Or 0.75 = 1.25k Or k = 3/5.

So AC:CB = K:1  = 3/5:1  Or 3:5.

The equation of the line AB as at (1) is y = 4(x-6)+5 should intersect the line x axis or y=0 at a point P(x' , 0) given by:

0=4(x'-6)+5 Or

4x' = 24-5 =19 Or

x' = 19/4 = 4.75

Therefore the interesecting point of the line AB and X axis  is P whose coordinates are (4.75,0).


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