SOLUTION: determine the value of k so that the line through points (0,4) and (k-2,6) has x-intercept k

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Question 1122351: determine the value of k so that the line through points (0,4) and (k-2,6) has x-intercept k
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
y=mx%2Bb
y=%28%286-4%29%2F%28k-2-0%29%29x%2B4, b=4given
y=%282%2F%28k-2%29%29x%2B4


The condition (k,0) is a point on the line
0=%282%2F%28k-2%29%29%2Ak%2B4

2k%2F%28k-2%29=-4

2k=-4%28k-2%29

2k=-4k%2B8
cross%28k=-2k%2B8%29k=-2k%2B4
3k=4
highlight%28k=4%2F3%29

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

Since the post by  @josgarithmetic  contains  ERRORS,  I came to fix them. 


y = mx + b

y = %28%286-4%29%2F%28k-2-0%29%29x%2B4,    b= 4  is given

y=%282%2F%28k-2%29%29x%2B4



According to the condition, the point  (k,0)  is a point on the line
0=%282%2F%28k-2%29%29%2Ak%2B4


2k%2F%28k-2%29 = -4


2k = -4(k-2)


2k = -4k + 8

k = -2k + 4

3k = 4


Answer.  k = 4%2F3.