# The points (2,-3), (3,-2) & (8,k) lie on a straight line. Find k.

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Let A(2,-3) B(3,-2) and C(8,k) lie on a straight line

Then, the slope for BC (m1) = the slope for AC (m2

Let us calculate:

m1= k+3)/(8-2) = (k+3)/6

m2 = (k+2)/(8-3) = (k+2)/5

m1 = m2

==> (k+3)/6 = (k+2)/5

Cross multiply:

==> 5(k+3) = 6(k+2)

==> 5k + 15 = 6k+ 12

==> 15-12 = 6k - 5k

==> k = 3

If the given points are located on the same line, the value of the determinant formed by the coordinates of 3 points has to be zero:

2 -3 1

determinant = 3 -2 1

8 k 1

We'll calculate the determinant:

det. = (2*-2*1) + 3*k*1 + 1*8*(-3) - (-2)*1*8 - 2*k - (-3)*3*1

det. = -4 + 3k - 24 + 16 - 2k + 9

But the determinant has to be equal to zero, if the points are on the line.

-4 + 3k - 24 + 16 - 2k + 9 = 0

We'll combine like terms:

k - 3 = 0

**k = 3**

We know that any 3 points (x1,y1) , (x2,y2) and (x3,y3) form a triangle whose area is {(1(y1+y2)(x2-x1)+(y2+y3)(x2-x3)+(y3+y1)}/2(x1-2).

So the given three points (2-3), (3,-2) and(8,k) should make a triangle of zero area to be on a line:

So the area of the given 3 points:

(-3-2)(2-3) +(-2+k)(3-8) +(k-3)(2-8) = 0

(-5)(-1 )+ (10-5k) +(-6k+18) = 0

5+10-5k-6k +18 = 0

33 -11k = 0

--11k = -33

k = -33/-11 = 3

k=3.

This is rise over run formula: y2-y1/x2-x1= slope

Since -2=y2 and -3=y1 and 3=x2 and 2=x1,

the slope m= -2-(-3)/3-2= 1

the formula for line is then y=x+n.

In order to find the n, you can substitute either (2,-3) or (3,-2).

I'll substitute the first one.

-3=2+n--> n=-5

Therefore, the line that the three points lie on is y=x-5.

You can substitute 8=x and k=y to find what k is.

y=x-5

k=8-5=3

k is 3.