Geometry help will give brainliest
1 answer:
Problem 1
=========
Write an equation for line passing through (8,12) that is perpendicular to
y = (4/3)x - 5
Note:
When two lines are perpendicular, the product of their slopes is -1.
The slope of the given line is 4/3.
Therefore the slope of the perpendicular line is -3/4.
Let the equation of the perpendicular line be
y = -(3/4)x + c
Because the line passes through (8,12), therefore
-(3/4)*8 + c = 12
-6 + c = 12
c = 18
The equation is y = -(3/4)x +18
Answer:
Problem 2
=========
Write an equation for a line passing through (-30,7) that is perpendicular to
y = -3x - 5.
The perpendicular line will have a slope of 1/3. Let its equation be
y = (1/3)x + c
Because the line passes through (-30,7), therefore
(1/3)*(-30) + c = 7
-10 + c = 7
c = 17
The equation is y = (1/3)x + 17
Answer:
=========
Write an equation for line passing through (8,12) that is perpendicular to
y = (4/3)x - 5
Note:
When two lines are perpendicular, the product of their slopes is -1.
The slope of the given line is 4/3.
Therefore the slope of the perpendicular line is -3/4.
Let the equation of the perpendicular line be
y = -(3/4)x + c
Because the line passes through (8,12), therefore
-(3/4)*8 + c = 12
-6 + c = 12
c = 18
The equation is y = -(3/4)x +18
Answer:
Problem 2
=========
Write an equation for a line passing through (-30,7) that is perpendicular to
y = -3x - 5.
The perpendicular line will have a slope of 1/3. Let its equation be
y = (1/3)x + c
Because the line passes through (-30,7), therefore
(1/3)*(-30) + c = 7
-10 + c = 7
c = 17
The equation is y = (1/3)x + 17
Answer:
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Answer
B)
Step by step explanation
The standard form of circle when center(h, k) and radius r, given is
(x - h)^2 + (y - k)^2 = r^2
Given: (h, k) = (5, 0) and r = 3
Now plug in these values in the standard form, we get
(x - 5)^2 + (y - 0)^2 = 3^2
Therefore, the answer is B)
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