<u>Answer:
</u>
The point-slope form of the line that passes through (6,1) and is parallel to a line with a slope of -3 is 3x + y – 19 = 0
<u>Solution:
</u>
The point slope form of the line that passes through the points 
and parallel to the line with slope “m” is given as
--- eqn 1
Where “m” is the slope of the line. 
are the points that passes through the line.
From question, given that slope “m” = -3
Given that the line passes through the points (6,1).Hence we get 
By substituting the values in eqn 1, we get the point slope form of the line which is parallel to the line having slope -3 can be found out.
y – 1 = -3(x – 6)
y – 1 = -3x +18
On rearranging the terms, we get
3x + y -1 – 18 = 0
3x + y – 19 = 0
Hence the point slope form of given line is 3x + y – 19 = 0
Area: Trapezoids A = ½h(b1<span> + b</span>2<span>)
</span>Rhombuses A=pq/2
Kites A=pq/2
To find the equation of this line in slope-intercept form (y = mx + b, where m is its slope and b is its y-intercept), we naturally need the slope and the y-intercept. We can see that the line intersects the y-axis at the point (0, 4) so our y-intercept is 4, and the line rises 4 along the y-axis for every 2 it runs along the x-axis, so its slope is 4/2 = 2. With this in mind, we can write the line's equation as
y = 2x + 4
3,162? I think it’s that I’m not sure.
Answer:
The area of the sector is 9.43 ft^2
Step-by-step explanation:
Here, we want to find the area of the sector with a central angle of 30 degrees
To do this, we use the formula below;
Area of sector = theta/360 * pi * r^2
theta = central angle = 30 degrees
r = radius = 6 ft
Thus, we have it that;
Area of sector = 30/360 * pi * 6^2
= 30/10 * pi = 3 * 3.142 = 9.43 ft^2
