# Thread: Find the Value of k

1. ## Find the Value of k

The vertices of Triangle PQR are P(1, 2), Q(-3, 6) and
R(4, 8). A line through Q is parallel to line segment PR and passes through point (k, 14). Find k.

MY WORK:

I know that parallel lines have equal slopes. I first found the slope of line segment PR to be 2/1.

I then equated the slope this way:

2/1 = (14 - 2)/(k - 1)

Solving for k, I got k = 7 but the answer is k = 1.

What did I do wrong and why is the answer 1 not 7?

2. Originally Posted by magentarita
The vertices of Triangle PQR are P(1, 2), Q(-3, 6) and
R(4, 8). A line through Q is parallel to line segment PR and passes through point (k, 14). Find k.

MY WORK:

I know that parallel lines have equal slopes. I first found the slope of line segment PR to be 2/1.

I then equated the slope this way:

2/1 = (14 - 2)/(k - 1)

Solving for k, I got k = 7 but the answer is k = 1.

What did I do wrong and why is the answer 1 not 7?
Hello Magentarita,

You used the wrong two points. You used point P instead of point Q.

Q(-3, 6) and X(k, 14)

Line QX will have a slope of 2 as you indicated.

$\displaystyle \frac{14-6}{k-^-3}=\frac{2}{1}$

This will get you to your desired result.

3. ## ok...........

Originally Posted by masters
Hello Magentarita,

You used the wrong two points. You used point P instead of point Q.

Q(-3, 6) and X(k, 14)

Line QX will have a slope of 2 as you indicated.

$\displaystyle \frac{14-6}{k-^-3}=\frac{2}{1}$

This will get you to your desired result.
I used the wrong points. No wonder I ended up with the wrong answer for k. I really enjoy geometry.