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Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3. - Mathematics

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Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Find the coordinates of the point which divides the join of A(–1, 7) and (4, –3) in the ratio 2 : 3.

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Solution 1

Let P(x, y) be the required point. Using the section formula

`x = (2xx4+3xx(-1))/(2+3) = (8-3)/5 = 5/5 = 1`

`y = (2xx(-3)+3xx7)/(2+3) =  (-6 + 21)/5 = 15/5 = 3`

Therefore the point is (1,3).

Solution 2

The end points of AB are  A(-1,7) and B (4,-3) 
Therefore `(x_1 = -1, y_1 = 7) and (x_2 = 4, y_2 = -3 )`

Also , m= 2 and n= 3
Let the required point be P (x ,y). 

By section formula, we get

`x= ((mx_2 + nx_1))/((m+n)) ,   y = ((my_2+ny_1))/((m+n))`

`⇒ x = ({ 2 xx 4 +3 xx (-1) })/(2+3) , y= ({2 xx (-3) + 3 xx 7})/(2+3)`

`⇒  x = (8-3) /5  , y = (-6+21)/5`

`⇒  x = 5/5 , y = 15/5`

Therefore, x = 1 and y= 3
Hence, the coordinates of the required point are (1,3) .

Concept: Section Formula
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APPEARS IN

NCERT Class 10 Maths
Chapter 7 Coordinate Geometry
Exercise 7.2 | Q 1 | Page 167
RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 1

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