use slopes to show that the triangle with vertices
(-2,7), (6,9) and (3,4) is a right triangle.
write an equation of the line that has y-intercept 5/7 and is parallel to the graph of 6x-3y=1
A(-2 , 7) , B(6 , 9).
m1 - (9 - 7) / (6 - (-2)) = 2 / 8 =1/4.
B(6 , 9) , C(3 , 4).
m2 = (4 - 9) / (3 - 6) = -5 / -3 = 5/3.
A(-2 , 7) , C(3 , 4).
m3 = (4 - 7) / (3 - (-2)) =-3 / 5=-3/5.
The slope of BC is the neg. reciprocal of AB. Therefore, the two lines are
perpindicular and form a rt triangle.
m = -A / B = -6 / -3 = 2.
Y = mx + b,
Y = 2x + 5 / 7.
To use slopes to show that a triangle is a right triangle, we need to find the slopes of its sides. If the product of the slopes of two sides is -1, then the triangle is a right triangle.
Let's find the slopes of the sides of the triangle with vertices (-2,7), (6,9), and (3,4).
The slope of a line passing through two points (x₁, y₁) and (x₂, y₂) is given by the formula:
m = (y₂ - y₁) / (x₂ - x₁)
For the side formed by (-2,7) and (6,9):
m₁ = (9 - 7) / (6 - (-2))
= 2 / 8
= 1/4
For the side formed by (6,9) and (3,4):
m₂ = (4 - 9) / (3 - 6)
= -5 / -3
= 5/3
For the side formed by (3,4) and (-2,7):
m₃ = (7 - 4) / (-2 - 3)
= 3 / -5
= -3/5
Now, let's check if the product of the slopes is -1:
(m₁) * (m₂) = (1/4) * (5/3) = 5/12
(m₂) * (m₃) = (5/3) * (-3/5) = -1
Since the product of the slopes of the sides formed by (6,9) and (3,4) and by (3,4) and (-2,7) is -1, we can conclude that the triangle is a right triangle.
Now, let's find the equation of the line that is parallel to the graph of 6x - 3y = 1 and has a y-intercept of 5/7.
The given equation is:
6x - 3y = 1
To find an equation of the line parallel to this, we need to keep the same slope. Let's rearrange the equation into slope-intercept form (y = mx + b), where m is the slope:
6x - 3y = 1
-3y = -6x + 1
y = 2x - 1/3
The slope of the given line is 2. Since a line parallel to this will have the same slope, the equation of the parallel line is:
y = 2x + b
Given that the y-intercept is 5/7, we can substitute the values into the equation:
y = 2x + b
5/7 = 2(0) + b
5/7 = b
Therefore, the equation of the line that has a y-intercept of 5/7 and is parallel to the graph of 6x - 3y = 1 is:
y = 2x + 5/7