Question

486) Find the Volue of k if given points Ote Collinear 1) (1,2), (-2,-10), (3,k)


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Answer 1

Given: Collinear points (1,2), (-2,-10), (3,k)

To find: value of K

Solution: Three or more points are said to be Collinear if they lie on a straight line.

Also we can say that for the three points to be Collinear, the area of the triangle must be zero.

That is, 1/2[ x1 ( y2 - y3) + x2 ( y3 - y1) + x3 ( y1 - y2)] = 0

x1 ( y2 - y3) + x2 ( y3 - y1) + x3 ( y1 - y2) = 0

here, (x1, y1) - (1,2)

(x2, y2) - (-2,-10)

(x3, y3) - (3,k)

by substituting these values in above equation;

1(-10-k)+ -2( k-2) + 3( 2-(-10))= 0

-10-k-2k+4+36=0

30= 3k

k= 10

Therefore the value of k for which the given points are Collinear will be k = 10

Answer 2

Answer:

The value of k is 10.

Step-by-step-explanation:

Let the three points be A, B and C.

• A ≡ ( 1, 2 ) ≡ ( x₁, y₁ )
• B ≡ ( - 2, - 10 ) ≡ ( x₂, y₂ )
• C ≡ ( 3, k ) ≡ ( x₃, y₃ )

We have given that,

The three points are collinear.

We know that,

If three points A, B, C are collinear, then the slope of lines AB & BC is equal.

$\displaystyle{\sf\:Slope\:of\:line\:AB\:=\:Slope\:of\:line\:BC}$

By slope formula,

$\displaystyle{\therefore\:\boxed{\pink{\sf\:\dfrac{y_2\:-\:y_1}{x_2\:-\:x_1}\:=\:\dfrac{y_3\:-\:y_2}{x_3\:-\:x_2}}}}$

$\displaystyle{\implies\sf\:\dfrac{-\:10\:-\:2}{-\:2\:-\:1}\:=\:\dfrac{k\:-\:(\:-\:10\:)}{3\:-\:(\:-\:2\:)}}$

$\displaystyle{\implies\sf\:\dfrac{\cancel{-\:12}}{\cancel{-\:3}}\:=\:\dfrac{k\:+\:10}{3\:+\:2}}$

$\displaystyle{\implies\sf\:4\:=\:\dfrac{k\:+\:10}{5}}$

$\displaystyle{\implies\sf\:k\:+\:10\:=\:4\:\times\:5}$

$\displaystyle{\implies\sf\:k\:+\:10\:=\:20}$

$\displaystyle{\implies\sf\:k\:=\:20\:-\:10}$

$\displaystyle{\implies\:\underline{\boxed{\red{\sf\:k\:=\:10\:}}}}$

∴ The value of k is 10.