Question

For
what value
value of P the
points (11,4), (1,-1) and (P, 1)
are collinear ?​

Answer 1

first we must find the equation with the points (11,4) and (1,-1)

y = mx + c

so,

slope:

$\frac{y2-y1}{x2-x1}$

which is

$\frac{4+1}{11-1} \\\\\frac{5}{10} = \frac{1}{2}$

now we know the slope is 1/4 we can substitute any point to get the y-intercept.

y intercept:

4 = 11/2 + c

8 = 11 + 2c

2c = 8 - 11

2c = -3

c = $\frac{-3}{2}$

equation: $y = \frac{1}{2} x -\frac{3}{2}$

now putting the point (P,1) in the equation:

$1 = \frac{P}{2} - \frac{3}{2}$

$1 = \frac{P-3}{2}$

2 = P -3

P = 2+3

P = 5

Answer:

P = 5