In an isosceles triangle, the perimeter is 8 more than 2 times on of the legs. If the perimeter is 28 in, find the length of the base.

A. 16 in
B. 18 in
C. 10 in
D. 8 in

Given triangle ABC with A(-3, 2), B(-1, -4), and C(4, 1), write the equation for the line containing altitude line AR in standard form.

A. x - y = 1
B. x + y = -1
C. x + y = 1
D. x - y = -1

Given triangle ABC with A(-3, 4), B(5, 8), and C(2, -2), write the equation of the line containing midsegment line XZ in standard form, where X is the midpoint of line AB and Z is the midpoint of line BC.

A. 6x - 5y = -24
B. 2x + 5y = 32
C. 2x - 5y = -28
D. 6x + 5y = 36

What is the image of O(-2, -1) after two reflections, first across the line y = -5, then across the line x = 1?

A. (-2, -1)
B. (-1, -6)
C. (4, -9)
D. (1, -5)

These are the only questions that I need help on for this test! I can't seem to solve them, I really need help!!! Please help? Thanks!

the amount of people who have seen this but havent said dip ship

Thanks Steve!

(But I STILL need help figuring out the AR one... gonna repost that question)

Sure, I'm here to help! Let's tackle these questions one by one.

1. In an isosceles triangle, the perimeter is 8 more than 2 times one of the legs. If the perimeter is 28 in, find the length of the base.
To solve this, let's assume the length of one of the legs is "x". Since it's an isosceles triangle, the other leg is also "x". The base of the triangle is what we need to find, so let's call it "y".

The perimeter is given by the equation: 2x + 2x + y = 28
Simplifying this equation gives us: 4x + y = 28.

Now, we know that the perimeter is also 8 more than 2 times one of the legs, which can be written as: 2x + 8.
Setting this equal to 28 gives us: 2x + 8 = 28.
Solving for x gives us: x = 10.

Plugging this value back into the equation 4x + y = 28, we get: 40 + y = 28.
Solving for y gives us: y = -12.

So, the length of the base is -12 in, which doesn't make sense for a triangle. Therefore, there seems to be an error in the problem statement.

2. The equation for the line containing the altitude line AR in standard form can be found using the coordinates of points A and R. However, we don't have the coordinates for point R, so we cannot determine the equation. Therefore, none of the answer choices (A, B, C, or D) are correct.

3. To write the equation of the line containing midsegment line XZ in standard form, we need to find the coordinates of point X and point Z.

Using the midpoint formula, we find that the coordinates of point X are (2, 6) (the average of the x-coordinates and the average of the y-coordinates of points A and B).

Similarly, the coordinates of point Z are (3.5, 3) (the average of the x-coordinates and the average of the y-coordinates of points B and C).

Now, we can use the point-slope formula to find the slope of the line containing XZ.
m = (y2 - y1) / (x2 - x1)
m = (3 - 6) / (3.5 - 2)
m = -3 / 1.5
m = -2

Plugging the slope and the coordinates of point X (2, 6) into the point-slope form equation, we get:
y - 6 = -2(x - 2)
y - 6 = -2x + 4
y = -2x + 10

To convert this into standard form, we move all terms to the left side of the equation:
2x + y = 10

Therefore, the correct answer is A. 2x + y = 10.

4. To find the image of point O(-2, -1) after two reflections, first across the line y = -5 and then across the line x = 1, we need to perform each reflection separately and then locate the final position of the point.

First Reflection: Across the line y = -5
The line y = -5 is a horizontal line, so the y-coordinate of point O remains the same. The x-coordinate will change sign, so O(-2, -1) becomes O(2, -1).

Second Reflection: Across the line x = 1
The line x = 1 is a vertical line, so the x-coordinate of point O remains the same. The y-coordinate will change sign, so O(2, -1) becomes O(2, 1).

Therefore, the final image of O(-2, -1) after two reflections is (2, 1). So, the correct answer is none of the choices (A, B, C, or D).

I hope this helps! Let me know if you have any further questions.

In any isosceles triangle, the perimeter is the base plus 2 times one of the legs. Clearly the base here is 8.

No idea where R is.

X = (A+B)/2 = (1,6)
Z = (B+C)/2 = (7/2,3)
XZ in two-point form is
y-6 = (-6/5)(x-1)
Now just rearrange that into standard form.

across y = -5: (x,y)->(x,-10-y)
across x = 1: (x,y)->(2-x,y)
both: (x,y)->(2-x,-10-y)
So, (-2,-1)->(4,-9)