Which equation is perpendicular to 3x - y = 9 and goes through the point (6,6)
1 answer:
Answer:
y = - x + 8
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Rearrange 3x - y = 9 into this form by subtracting 3x from both sides
- y = - 3x + 9 ( divide all terms by - 1 )
y = 3x - 9 ← in slope- intercept form
with slope m = 3
Given a line with slope m then the slope of a line perpendicular to it is
= -
= -
, thus
y = - x + c ← is the partial equation
To find c substitute (6, 6) into the partial equation
6 = - 2 + c ⇒ c = 6 + 2 = 8
y = - x + 8 ← equation of perpendicular line
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Let's first rewrite it in vertex form.
Start by completing the square.
y = x² + 4x + 4 - 4 - 1
y = (x + 2)² - 5
(x + 2)² = y + 5
Now, 4a = 1; a =
So, the vertex form is (x + 2)² = 4()(y + 5)
So, we know that the vertex is at (-2, -5).
Since it's a concave up parabola, we just have to change the y-value to find the focus.
∴ F(-2, -5 +)
F(-2,
)
Start by completing the square.
y = x² + 4x + 4 - 4 - 1
y = (x + 2)² - 5
(x + 2)² = y + 5
Now, 4a = 1; a =
So, the vertex form is (x + 2)² = 4(
So, we know that the vertex is at (-2, -5).
Since it's a concave up parabola, we just have to change the y-value to find the focus.
∴ F(-2, -5 +
F(-2,