How to solve k and this other question.

**dizzycarly** · June 21st 2009, 02:16 PM

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!

**yeongil** · June 21st 2009, 02:27 PM

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:
$\begin{aligned}<br /> 0 &= 4(-20) + k \\<br /> 0 &= -80 + k \\<br /> 80 &= k<br /> \end{aligned}$

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

01

**dizzycarly** · June 21st 2009, 06:22 PM

thanks very much

I forgot to say x.

does anyone know how to solve the second question?

**Jhevon** · June 21st 2009, 06:34 PM

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 $k$ ?

**VonNemo19** · June 21st 2009, 06:37 PM

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.

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