Math Help - How to solve k and this other question.

1. How to solve k and this other question.

Hey, can you guys tell me how to figure out this.

The graph of y=4x+k has a intercept of -20. Determine the value of k.

and.. how do you solve..

The line segiment joining R(k,6) and S(-4,k) has a midpoint at (-11,-6). Determine the value of k.

thanks!

2. Originally Posted by dizzycarly
Hey, can you guys tell me how to figure out this.

The graph of y=4x+k has a intercept of -20. Determine the value of k.
Which intercept? An x-intercept or a y-intercept?

If -20 is the y-intercept, then that's k. Slope-intercept form of the equation of the line is y = mx + b, and b is the y-intercept. So the equation is just
$y = 4x - 20$.

But if -20 is the x-intercept, then you plug in -20 for x and 0 for y:
\begin{aligned}
0 &= 4(-20) + k \\
0 &= -80 + k \\
80 &= k
\end{aligned}

So the equation would be
$y = 4x + 80$.

01

3. thanks very much I forgot to say x.

does anyone know how to solve the second question?

4. Originally Posted by dizzycarly
thanks very much I forgot to say x.

does anyone know how to solve the second question?
I like your avatar.

for the second problem, we have by the midpoint formula:

$(-11,-6) = \left( \frac {k - 4}2, \frac {6 + k}2 \right)$

now do you see how to solve for $k$?

5. Originally Posted by dizzycarly
thanks very much I forgot to say x.

does anyone know how to solve the second question?

If the point given in the problem is in fact the midpoint, it must satisfy these conditions

$(-11,-6) = \left( \frac {x_1+x_2}2, \frac {y_1+y_2}2 \right)$

therefore, let $x_1=k$ and $x_2=-4$

Then

$\frac{k+(-4)}{2}=-11$

Solve for k.

You can verify the y ordinate.