keeping in mind that perpendicular lines have <u>negative reciprocal</u> slopes, let's find the slope of 3x + 4y = 9, by simply putting it in slope-intercept form.
so we're really looking for the equation of a line whose slope is 4/3 and runs through 8, -4.
Answer:
A non-equilateral rhombus.
Step-by-step explanation:
We can solve this graphically.
We start with square:
ABCD
with:
A = (11, - 7)
B = (9, - 4)
C = (11, - 1)
D = (13, - 4)
Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:
AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13
BC = √( (11 - 9)^2 + (-1 -(-4))^2) = √13
CD = √( (11 - 13)^2 + (-1 -(-4))^2) = √13
DA = √( (11 - 13)^2 + (-7 -(-4))^2) = √13
And we change C by C' = (11, 1)
In the image you can see the 5 points and the figure that they make:
The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.
