girlworld_club
Member
If points (3k, 1) (k -1, k-3) and (k-4, k-5) are collinear (have the same gradient) find K.
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A( 3k,1) , B (k-1, k-3) & C (k-4, k-5)If points (3k, 1) (k -1, k-3) and (k-4, k-5) are collinear (have the same gradient) find K.
this solution is correctA( 3k,1) , B (k-1, k-3) & C (k-4, k-5)
m AB = [ (k-3) - 1 ] / [(k-1) -3k] = [ k-4 ] / [ -2k-1]
m BC = [ (k-5) - (k-3) ] / [ (k-4) - (k-1)] = [ -2 ] / [ -3] = 2/3
m AB = m BC
[k-4]/[-2k -1] = 2/3
3(k-4) = 2(-2k-1)
3k -12 = -4k -2
7k = 10
k = (10/7)
And you do MX2?If points (3k, 1) (k -1, k-3) and (k-4, k-5) are collinear (have the same gradient) find K.
Ill give you a latex versionit's usually the basic stuff that end up getting to me.
lol are your subjects legit, not referring to your ability not to do this questioni get it, it's how i originally solved the question but did a silly sbstitution mistake. Just a header, when putting an = sign always use a new line, i though the = represented that for example x=y (meaning they should be the same). and not that after solving x you should get y