Question

The value of k for which the points A (0,1),B(2,k) and C(4,-5)are collinear

Answer 1

THE VALUE OF K OF THE GIVEN POINTS IS -2

Answer 2

The value of k is -2

Step-by-step explanation:

here, given points are  A (0,1),B(2,k) and C(4,-5)

since , The points are collinear therefore , the area of triangle is equal to 0

then

$x_1 = 0 , y_1= 1 \\\\x_2=2 , y_2=k\\\\x_3=4 , y_3=-5$

area of triangle = $\frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))$

placing all the values

$\frac{1}{2}(0(k+5)+2(-5-1)+4(1-k))=0\\\\\frac{1}{2}(-12+4-4k)=0\\\\\frac{1}{2}(-8-4k)=0\\\\-8-4k=0\\\\k=-2$

hence , The value of k is -2

#Learn more:

Find the value of k for which the points A(k, 5) B(0,1)and C(2,-3) are collinear

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