# I know how to get the answer to this question, but I want to know if there's an algebraic way to solve it, with proportions. (I always have trouble with proportions.) Please use math symbols so I...

I know how to get the answer to this question, but I want to know if there's an algebraic way to solve it, with proportions. (I always have trouble with proportions.)

Please use math symbols so I can understand how to do this easily, too.

Thank you!

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gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

For division of line segment AB in ratio 5:2 by point C, the coordinates of point C can be calculated as

Xc = (5*Xb + 2*Xa)/(5+2) = (5*-7 + 2*0)/7 = -35/7 = -5.

Similarly, Yc = (5*Yb + 2*Ya)/(5+2). = (5*-4 + 2*10)/7 = 0.

Thus the point C is (-5,0).

Hope this helps.

kspcr111 | In Training Educator

Posted on

Hi your answer is ready ..... but it has got some explanation to do so i had to write it on a paper and upload it for you

plz check the attachment for detailed solution.

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kspcr111 | In Training Educator

Posted on

Using the proportions i tried to solve this ....plz check hope this might help you .

Let, (x, y) be the required co-ordinates of C. From A,B,C Draw AL, BM and CN perpendiculars on OX. Again, draw AT parallel to OX to cut CN at S and BM and CN at S and T respectively, Then,

AS = LN = ON - OL = x – x1;

AT = LM = OM – OL = x2 - x1;

CS = CN – SN = CN – AL = y - y1;

and BT = BM – TM = BM – AL = y2 – y1

Again, AC/CB = m/n

or, CB/AC = n/m

or, CB/AC + 1 = n/m + 1

or, ((CB + AC)/AC) = (m + n)/m

o, AB/AC = (m + n)/m

Now, by construction, the triangles ACS and ABT are similar; hence,

AS/AT = CS/BT = AC/AB

Taking, AS/AT = AC/AB

we get,

(x - x1)/(x2 - x1) = m/(m + n)

or, x (m + n) – x1 (m + n) = mx2 – mx1

or, x ( m + n) = m*x2 - m*x1 + m*x1 + n*x1 = m*x2 + n*x1

Therefore, x = (m*x2 + n*x1)/(m + n)
Again, taking CS/BT = AC/AB we get,

(y - y1)/(y2 - y1) = m/(m + n)

or, ( m + n) y - ( m + n) y1 = m*y2 – m*y1

or, ( m+ n)y = m*y2 – m*y1 + m*y1 + n*y1 = m*y2 + n*y1

Therefore, y = (m*y2 + n*y1)/(m + n)

Therefore, the required co-ordinates of the point R are
[(m*x2 + n*x1)/(m + n), (m*y2 + n*y1)/(m + n)]------------------(1)

Now in the problem

(x1,y1)=(0,10)

(x2,y2)=(-7,-4)

m:n = 5:2

By using equation (1) we get,

C(X,Y) = [5*(-7)+2(0)/7 , 5*(-4)+2(10)/7  ]

= [-5,0]

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balajia | College Teacher | (Level 1) eNoter

Posted on

Any point C(x,y) which divides the line joining A(x1,y1) and B(x2,y2) in the ratio m:n

then c(x,y)=((mx2+nx1)/m+n, (my2+ny1)/m+n)

here in the given question A(0,10) and B(-7,-4)

and m=5 and n=2

substitute in the above formula then you will get C(-5,0)