Analytic Geometry Q6

• Jan 31st 2009, 02:13 AM
looi76
Analytic Geometry Q6
Question:
Using slopes, determine if the points lie on a straight line.
a) $(-1,-4)$ , $(3,8)$ & $(0,-1)$
b) $(0,5)$ , $(-1,3)$ & $(-3,2)$

Attempt:

a) $(-1,-4)$ & $(3,8)$

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8-(-4)}{3-(-1)} = \frac{12}{4} = 3$

$(3,8)$ & $(0,-1)$

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1-8}{0-3} = \frac{-9}{-3} = 3$

It is a straight line as the slopes are the same.

b) $(0,5)$ & $(-1,3)$

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3-5}{-1-0} = 2$

$(-3,2)$ & $(-1,3)$

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3-2}{(-1)-3} = \frac{1}{-4} = -\frac{1}{4}$

It is not a straight line as the slopes don't match.

• Jan 31st 2009, 02:27 AM
running-gag
a) you data is (0,1) but you made the calculation with (0,-1)

b) your calculation of the second slope is not correct
• Jan 31st 2009, 02:35 AM
looi76
a) I have corrected the question. Is my answer right?

b) $(-3,2)$ & $(-1,3)$

$\color{red}{m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3-2}{(-1)-3} = \frac{1}{-4} = -\frac{1}{4}}$

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2-3}{-3-1} = \frac{1}{4}$

It is still not a straight line, am I right?
• Jan 31st 2009, 04:14 AM
running-gag
a) OK

b) The slope is still not correct ...