Question

It the points (7.-2] [5.1].[3.k] Jore
collinear find the volue
of 'k!​

Answer 1

Step-by-step explanation:

Given :-

the points (7,-2), (5,1) and (3, k) are collinear

To find :-

Find the value of k?

Solution:-

Given points are (7,-2), (5,1) and (3, k)

Let (x1, y1)=(7,-2)=>x1=7 and y1 = -2

Let (x2, y2)=(5,1)=>x2=5 and y2=1

Let (x3, y3)=(3,k)=>x3=3 and y3=k

Given that

The points are Collinear points

=> The area of the triangle formed by the given points is equal to 0

We know that

The area of the triangle formed by the points

(x1, y1) ,(x2, y2) and (x3, y3) is

(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) |

We have

(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | = 0

=> (1/2) | 7(1-k)+5(k+2)+3(-2-1) | = 0

=> (1/2) | 7(1-k) +5(k+2)+3(-3) | = 0

=> (1/2) | 7-7k+5k+10-9 | = 0

=> (1/2) | 8-2k | = 0

=> | 8-2k | = 0×(2/1)

=> | 8-2k | = 0

=> 8-2k = 0

=> 8 = 2k

=> 2k = 8

=> k = 8/2

=> k = 4

Therefore ,k = 4

Answer:-

The value of k for the given problem is 4

Used formulae:-

Collinear Points:-

The points lie on the same line are called Collinear Points.

If the points are Collinear then the area of the triangle formed by the given points is equal to 0.

Area of a triangle :-

The area of the triangle formed by the points

(x1, y1) ,(x2, y2) and (x3, y3) is ∆=

(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units