# Math Help - lines

1. ## lines

Need help yet again.

1) The line 3x-4y=4 is midway between the lines l1 and l2, such that l1 and l2 are 8 units perpendicular from each other. Find the equations of lines l1 and l2, express in general form.

2) Find the value of k so that the circle whose equation (x-6)^2 + (y-3)^2 = 20 is tangent to the line 2x-y+k=0.

3) The sides of a triangle are located on the following linear equations: 5x+y-16=0; and x+4y=26. Find the vertices of the triangle, as well as area and perimeter.

4) Find the equation of the tangent line to the circle whose equation is x^2+y^2-4x-6y+5=0, and this tangent line is perp. to the line x-y=7.

5) Find the equation of a circle tangent to the line x+y=8 at the point (2,6) and passing through the point (4,0)

2. Hello, archistrategos214!

These are not simple problems . . . Here's #2.

2) Find the values of $k$ so that the circle $(x -6)^2+(y-3)^2\:=\:20$
is tangent to the line $2x-y +k\:=\:0$

We will find the intersections of the line and the circle.

We have: . $\begin{array}{cc}y \:=\:2x + k & (1)\\(x-6)^2+(y-3)^2\:=\:20 & (2) \end{array}$

Substitute (1) into (2): . $(x-6)^2 + (2x+k-3)^2\:=\:20$

. . and we have the quadratic: . $5x^2 + 4(k-6)x + (k^2-6k+25) \:=\:0$

Quadratic Formula: . $x \;=\;\frac{-4(k-6) \pm \sqrt{16(k-6)^2 - 4(5)(k^2-6k+25)}}{2(5)}$

. . which simplifies to: . $x \;=\;\frac{-4(k-6) \pm\sqrt{76 - 72k - 4k^2}}{10}$

The line and circle are tangent if there is one point of intersection.
. . This occurs when the discriminant is zero.

We have: . $-4k^2 - 72k + 76 \:=\:0\quad\Rightarrow\quad -4(k - 1)(k + 19) \:=\:0$

Therefore: . $\boxed{k \:=\:1,\:-19}$

3. Thanks Soroban!

Anybody care to help with number 3? It's killing me, lol. Anyway, thanks for all the help! Couldn't have done it otherwise

4. Hello again, archistrategos214!

3) The sides of a triangle are located on the following linear equations:
. . $5x+y-16\:=\:0$ and $x+4y\:=\:26$

Find the vertices of the triangle, as well as area and perimeter.

You gave us only two sides . . . We don't have a triangle.

5. Originally Posted by archistrategos214
Need help yet again.
1) The line 3x-4y=4 is midway between the lines l1 and l2, such that l1 and l2 are 8 units perpendicular from each other. Find the equations of lines l1 and l2, express in general form.
In this problem is would be usefule to have 2 form for the line:
"Standard Form"
$3x-4y-4=0$
"Slope-Intercept Form"
$y=\frac{3}{4}x-1$.

Accoring to the problem there are two parrallel line $l_1,l_2$ such that one is "above" and one is "below". We are also told the distance between them is 8. That means the distance between $l_1,l_2$ and the line we are given is 4.
Since the lines are parrallel the distance can be computed as easily. First take any point on the line and compute its distance (from the distane formula to a line) to the other parallel line (In this case 4).
Since $l_1$ is parralel with $y=\frac{3}{4}x-12$ it must have the same slope, just a different intercept. Thus,
$y=\frac{3}{4}x+k$. Where $k$ is the value that makes the distance between them 4. Take any point on this line, say $x=0$ then $y=k$. Thus, $(0,k)$ is on the line $l_1$. Now find the distance from this point to the other line, we use the formula:
$\boxed{ \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}}$.
Remember in the beginning we found the standard form of the line, which was
$3x-4y-4=0$
By the formula we need to evaluate this at $(0,k)$.
Thus,
$3(0)-4k-4=-4k-4$.
And we need to find,
$\sqrt{A^2+B^2}$
Which is,
$\sqrt{3^2+(-4)^2}=5$.
Thus, by the formula and the problem says distance is 4 we have,
$\frac{|-4k-4|}{5}=4$
Multiply by 5,
$|-4k-4|=20$
This is an absolute value equation, two solutions,
$-4k-4=\pm 20$
Add 4 and divide by -4,
$k=-1\mp 5=4,-6$
These values gives us the two intercepts for the two lines.
Thus one line is,
$y=\frac{3}{4}x+4$
And the other,
$y=\frac{3}{4}x-6$

6. Hello, archistrategos214!

There's surely an easier way to do #5 . . . but here's my solution.

5) Find the equation of a circle tangent to the line $x+y\:=\:8$ at the point $(2,6)$
and passing through the point $(4,0)$.
Code:
          \ |
*    A
| \(2,6)
|   *
| /   \
/       \
/ |         \
/   |           \
C *     |             \
- - - - - + - - - * - - - * - -
|     (4,0)       \
|       B

The center of the circle is $C$.

Since the circle is tangent to the line at point $A$,
. . the radius at $A$ is perpendicular to the line.
The center $C$ lies along this perpendicular.

The line is: . $y \:=\:-x + 8$ with slope $-1$.
The equation of the perpendicular is: . $y - 6 \:=\:1(x - 2)\quad\Rightarrow\quad y\:=\:x + 4$ [1]

The center $C$ also lies on the perpendicular bisector of chord $AB$.

The midpoint of $AB$ is: $(3,3)$

The slope of $AB\:=\:\frac{0-6}{4-2}\:=\:-3$
. . The slope of the perpendicular bisector is: $+\frac{1}{3}$
The equation of the bisector is: . $y - 3 \:=\:\frac{1}{3}(x - 3)\quad\Rightarrow\quad y \:=\:\frac{1}{3}x + 2$ [2]

The center is the intersection of [1] and [2]: . $x + 4 \:=\:\frac{1}{3}x + 2\quad\Rightarrow\quad x = -3,\;y = 1$
. . Hence, the center is: $C(-3,1)$

The radius is the length of $CA:\;\;r^2 \:=\:(2+3)^2 + (6-1)^2 \:=\:50$

Therefore, the equation of the circle is: . $\boxed{(x+3)^2 + (y-1)^2\:=\:50}$

7. Originally Posted by Soroban
Hello again, archistrategos214!

You gave us only two sides . . . We don't have a triangle.

Hello Soroban,

Whoops, sorry about that. The three sides are as follows: 5x+y-16=0; 4x-3y-28=0; and x+4y=26.

8. Originally Posted by archistrategos214
Hello Soroban,

Whoops, sorry about that. The three sides are as follows: 5x+y-16=0; 4x-3y-28=0; and x+4y=26.
You need to find where the lines intersect, such as where does line 1 cross line 2, line 3 cross line 1, and line 2 cross line 3?

Then apply the distance formula: $d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

Do that three times, and you'll find the perimeter.

As for area, do you know Heron's formula?

9. Originally Posted by Quick
You need to find where the lines intersect, such as where does line 1 cross line 2, line 3 cross line 1, and line 2 cross line 3?

Then apply the distance formula: $d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

Do that three times, and you'll find the perimeter.

As for area, do you know Heron's formula?
I'm pretty certain I have Heron's formula somewhere in my notes, I just need to be able to find it. And thanks for the tip

10. Hello, archistrategos214!

3) The sides of a triangle are located on the following linear equations:
. . $\begin{array}{ccc} 5x+y\:=\:16 \\ x+4y\:=\:28 \\ x + 4y \:=\:26\end{array}$

Find the vertices of the triangle, as well as area and perimeter.

The vertices are the intersections of the three lines, taken in pairs.

We have: . $\begin{array}{ccc}(A)\\(B)\\(C)\end{array}\begin{a rray}{ccc}5x + y \:=\:16 \\ 4x - 3y \:=\:28 \\ x + 4y \:=\:26\end{array}$

$\begin{array}{cc}3\times (A) & 15x + 3y \:=\:48 \\ \text{Add }(B) & 4x - 3y \:=\:28\end{array}\quad\Rightarrow\quad 19x = 76\quad\Rightarrow\quad x = 4,\:y = \text{-}4$

Lines (A) and (B) intersect at point $P(4,\,\text{-}4)$

$\begin{array}{cc}\text{-}4\times(C) & \text{-}4x - 16y \:=\:\text{-}104 \\ \text{Add (B)} & 4x - 3y \:=\:28\end{array}\quad\Rightarrow\quad \text{-}19y =\text{-}76\quad\Rightarrow\quad y = 4,\:x = 10$

Lines (B) and (C) intersect at point $Q(10,4)$

$\begin{array}{cc}\text{-}4\times(A) & \text{-}20x - 4y \:=\:\text{-}64\\ \text{Add (C)} & x + 4y \:=\:26\end{array}\quad\Rightarrow\quad \text{-}19x = \text{-}38\quad\Rightarrow\quad x = 2,\;y = 6$

Lines (A) and (C) intersect at point $R(2,6)$

We have the triangle with vertices: . $\boxed{P(4,\text{-}4),\;Q(10,4),\;R(2,6)}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The lengths of the sides are:

. . $PQ \:=\:\sqrt{(10-4)^2 + (4-[\text{-}4])^2}\:=\:\sqrt{6^2+8^2} \:=\:\sqrt{100} \:=\:10$

. . $QR \:=\:\sqrt{(2-10^2 + (6-4)^2} \:=\:\sqrt{8^2 + 2^2}\:=\:\sqrt{68} \:=\:2\sqrt{17}$

. . $PR \:=\:\sqrt{92-4)^2 + (6-[\text{-}4])^2} \:=\:\sqrt{2^2+10^2} \:=\:\sqrt{104}\:=\:2\sqrt{26}$

The perimeter is: . $\boxed{10 + 2\sqrt{17} + 2\sqrt{26}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The area of a triangle with vertices $(x_1,y_1),\:(x_2,y_2),\:(x_3,y_3)$

. . is given by: . $\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix}$

We have: . $\frac{1}{2}\begin{vmatrix}4 & \text{-}4 & 1 \\ 10 & 4 & 1 \\ 2 & 6 & 1\end{vmatrix}\;=\;\frac{1}{2}(76)\quad\Rightarrow \quad \boxed{\text{Area } = \:38}$