Question

Find the value of k , if (k,3) , ( 6,-2) and (-3,4) are collinear

Answer 1

To prove ;

Given points = collinear

OR

ar(∆ABC) = 0

Given;

• A(k,3)
• B(6,-2)
• C(-3,4)

Thus area of ∆ in coordinate geometry ;

$\frac{1}{2} |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) \: |$

Or for the given situation ;

$|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) \: | = 0$

• This is because if a ∆ has its area = 0 then all 3 vertices of it will occur in the same straight line and so collinear

• k(-2-4) + 6(4-3) -3(3+2) = 0
• -6k + 6(1) -3(5) = 0
• -6k + 6 - 15 = 0
• -6k - 9 = 0
• -6k = 9
• 6k = -9

Then;

$k \: = \frac{ - 9}{6} \\ \\ k \: = \frac{ - 3}{2} \\ \\ thus \: k \: = \frac{ - 3}{2} or \: - 1.5$

#answerwithquality

#BAL