# equation of a straight line.

• February 13th 2009, 01:03 PM
Tweety
equation of a straight line.
The straight line passing through the point P(2,1) and the point Q(k,11) has gradient $- \frac{5}{12}$

(a) Find the equation of the line in terms of x and y only.
(b) Determine the value of k

Not sure how to go about doing this question.
• February 13th 2009, 01:29 PM
masters
Quote:

Originally Posted by Tweety
The straight line passing through the point P(2,1) and the point Q(k,11) has gradient $- \frac{5}{12}$

(a) Find the equation of the line in terms of x and y only.
(b) Determine the value of k

Not sure how to go about doing this question.

Hi Tweety,

Let's use the point-slope form of the linear equation: $y-y_1=m(x-x_1)$.

$y-1=-\frac{5}{12}(x-2)$

If you have to convert this equation to slope-intercept or standard form, I'll assume you can handle that task.

As far as finding k, let's set up the slope formula with k in it like this:

$m=\frac{y_1-y_2}{x_1-x_2}$

$\frac{-5}{12}=\frac{11-1}{k-2}$

$\frac{-5}{12}=\frac{10}{k-2}$

$-5(k-2)=120$

$k-2=-24$

$k=-22$