Points A(–6, 10), B(–4, 6) and C (3, –8) are collinear such that AB = 29 AC. - Mathematics

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Points A(–6, 10), B(–4, 6) and C(3, –8) are collinear such that AB = `2/9` AC.

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  • True

  • False

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Solution

This statement is True.

Explanation:

If the area of triangle formed by the points (x1, y2), (x2, y2) and (x3, y3) is zero, then the points are collinear,

∵ Area of triangle = `1/2[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]`

Here, x1 = – 6, x2 = – 4, x3 = 3 and y1 = 10, y2 = 6, y3 = – 8

∴ Area of ΔABC = `1/2[-6{6 - (-8)} + (-4)(-8 - 10) + 3(10 - 6)]`

= `1/2[-6(14) + (-4)(-18) + 3(4)]`

= `1/2(-84 + 72 + 12)`

= 0

So, given points are collinear.

Now, distance between A(– 6, 10), B(– 4, 6), 

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

AB = `sqrt((-4 + 6)^2 + (6 - 10)^2`

`sqrt(2^2 + 4^2)`

= `sqrt(4 + 16)`

= `sqrt(20)`

= `2sqrt(5)`

Distance between A(– 6, 10) and C(3, – 8),

AC = `sqrt((3 + 6)^2 + (-8 - 10)^2`

= `sqrt(9^2 + 18^2)`

= `sqrt(81 + 324)`

= `sqrt(405)`

= `sqrt(81 xx 5)`

= `9sqrt(5)`

∴ AB = `2/9` AC

Which is the required relation.

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Chapter 7: Coordinate Geometry - Exercise 7.2 [Page 81]

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NCERT Exemplar Mathematics Class 10
Chapter 7 Coordinate Geometry
Exercise 7.2 | Q 10 | Page 81

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