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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 9

The line segment joining A(6, 3) and B(−1, −4) is doubled in length by adding half of AB to each end. Find the coordinates of the new end points. - Mathematics

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Question

The line segment joining A(6, 3) and B(−1, −4) is doubled in length by adding half of AB to each end. Find the coordinates of the new end points.

Sum

Solution


Distance = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt((6 + 1)^2 + (3 + 4)^2`

= `sqrt(7^2 + 7^2)`

= `sqrt(49 + 49)`

= `7sqrt(2)`

By given, condition

AA' = `(7sqrt(2))/2`, BB' = `(7sqrt(2))/2`


`"BA'"/"AA'" = (7sqrt(2) + (7sqrt(2))/2)/((7sqrt(2))/2)`

= `(14sqrt(2) + 7sqrt(2))/2 ÷ (7sqrt(2))/2`

= `(21sqrt(2))/2 xx 2/(7sqrt(2))`

= `3/1`

m : n = 3 : 1

A line divides externally in the ratio m : n

The point P is `(("m"x_2 - "n"x_1)/("m" - "n"), ("m"y_2 - "n"y_1)/("m" - "n"))`

m = 3, n = 1, x1 = −1, x2 = 6, y1 = 4, y2 = 3

The Point A' = `((18 + 1)/(3 - 1), (9 + 4)/(3 - 1))`

= `(19/2, 13/2)`

m = 3, n = 1, x1 = 6, x2 = −1, y1 = 3, y2 = 4

The Point B' = `((-3 - 6)/(3 - 1), (-12 - 3)/(3 - 1))`

= `((-9)/2, (-15)/2)`

The new end points are `(19/2, 13/2)` and `((-9)/2, (-15)/2)`

Aliter:

Distance = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt((6 + 1)^2 + (3 + 4)^2`

= `sqrt(7^2 + 7^2)`

= `sqrt(49 + 49)`

= `7sqrt(2)`

AA' = `-7sqrt(2)`  ...(add half of AB)

 = `(7sqrt(2))/2`

`"AB"/"AA'" = 7sqrt(2) ÷ 7sqrt(2)/2`

= `(7sqrt(2) xx 2)/(7sqrt(2))`

= `2/1`

∴ BA’ divides in the ratio 2 : 1

A line divides internally in the ratio m : n the point is`("m"x_2 + "n"x_1)/("m" + "n"), ("m"y_2 + "n"y_1)/("m" + "n")`

Let the point A’ be (a, b)

(6, 3) = `((2"a" - 1)/(2 + 1), (2"b" - 4)/(2 + 1))`

= `((2"a" - 1)/3, (2"b" - 4)/3)`

`(2"a" - 1)/3` = 6

2a – 1 = 18

2a = 19

a = `19/2`

and

`(2"b" - 4)/3` = 3

2b – 4 = 9

2b = 13

b = `13/2`

The point A’ is `(19/2, 13/2)`

To find B’

Let B’ be (a, b)

AB = `7sqrt(2)`

BB' = `1/2 xx 7sqrt(2) = (7sqrt(2))/2`

`"AB"/"BB'" = 7sqrt(2) ÷ (7sqrt(2))/2`

= `(7sqrt(2) xx 2)/(7sqrt(2))`

= 2

AB’ divides in the ratio 2 : 1

(−1, −4) = `(2"a" + 6)/3, (2"b" + 3)/3`

`(2"a" + 6)/3` = −1

2a + 6 = −3

2a = −3 – 6

2a = −9

a = `-9/2`

and

`(2"b" + 3)/3` = −4

2b + 3 = −12

2b = −12 – 3

2b = −15

b = `-15/2`

= −1

The point B' is `(-9/2, -15/2)`

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Points of Trisection of a Line Segment (Mid-point Formula)
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Chapter 5: Coordinate Geometry - Exercise 5.4 [Page 214]

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Samacheer Kalvi Mathematics [English] Class 9 TN Board
Chapter 5 Coordinate Geometry
Exercise 5.4 | Q 5 | Page 214
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