Using point-slope formula for calculating equation of the line;
y - y1 = m (x – x1)
Where x1 and y1 is the point (0,2), i.e. x1 = 0 and y1 = 2
and slope = m = 2/3
Putting all values;
y – 2 = 2/3 (x – 0)
y – 2 = 2/3x
3y – 6 = 2x --------------- (1)
Now put (-3,0) in equation (1), i.e. x = -3 and y = 0;
– 6 = – 6
Hence (-3,0) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.
Now checking other points, whether it satisfies the equation of line or not.
putting (-2,-3) in equation (1);
- 9 - 6 = - 4
(-2,-3) does not lie on the line.
Putting (2,5) in equation (1);
15 – 6 = 4
(2,5) does not lie on the line.
Putting (3,4) in equation (1);
12 – 6 = 6
6 = 6
Hence (3,4) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.
Putting (6,6) in equation (1);
18 – 6 = 12
12 = 12
Hence (6,6) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.
Using point-slope formula for calculating equation of the line;
y - y1 = m (x – x1)
Where x1 and y1 is the point (0,2), i.e. x1 = 0 and y1 = 2
and slope = m = 2/3
Putting all values;
y – 2 = 2/3 (x – 0)
y – 2 = 2/3x
3y – 6 = 2x --------------- (1)
Now put (-3,0) in equation (1), i.e. x = -3 and y = 0;
– 6 = – 6
Hence (-3,0) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.
Now checking other points, whether it satisfies the equation of line or not.
putting (-2,-3) in equation (1);
- 9 - 6 = - 4
(-2,-3) does not lie on the line.
Putting (2,5) in equation (1);
15 – 6 = 4
(2,5) does not lie on the line.
Putting (3,4) in equation (1);
12 – 6 = 6
6 = 6
Hence (3,4) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.
Putting (6,6) in equation (1);
18 – 6 = 12
12 = 12
Hence (6,6) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.Using point-slope formula for calculating equation of the line;
y - y1 = m (x – x1)
Where x1 and y1 is the point (0,2), i.e. x1 = 0 and y1 = 2
and slope = m = 2/3
Putting all values;
y – 2 = 2/3 (x – 0)
y – 2 = 2/3x
3y – 6 = 2x --------------- (1)
Now put (-3,0) in equation (1), i.e. x = -3 and y = 0;
– 6 = – 6
Hence (-3,0) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.
Now checking other points, whether it satisfies the equation of line or not.
putting (-2,-3) in equation (1);
- 9 - 6 = - 4
(-2,-3) does not lie on the line.
Putting (2,5) in equation (1);
15 – 6 = 4
(2,5) does not lie on the line.
Putting (3,4) in equation (1);
12 – 6 = 6
6 = 6
Hence (3,4) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.
Putting (6,6) in equation (1);
18 – 6 = 12
12 = 12
Hence (6,6) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.Using point-slope formula for calculating equation of the line;
y - y1 = m (x – x1)
Where x1 and y1 is the point (0,2), i.e. x1 = 0 and y1 = 2
and slope = m = 2/3
Putting all values;
y – 2 = 2/3 (x – 0)
y – 2 = 2/3x
3y – 6 = 2x --------------- (1)
Now put (-3,0) in equation (1), i.e. x = -3 and y = 0;
– 6 = – 6
Hence (-3,0) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.
Now checking other points, whether it satisfies the equation of line or not.
putting (-2,-3) in equation (1);
- 9 - 6 = - 4
(-2,-3) does not lie on the line.
Putting (2,5) in equation (1);
15 – 6 = 4
(2,5) does not lie on the line.
Putting (3,4) in equation (1);
12 – 6 = 6
6 = 6
Hence (3,4) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.
Putting (6,6) in equation (1);
18 – 6 = 12
12 = 12
Hence (6,6) lie on the line, as it satisfies the equation of line, so vera can use (-3,0) to graph the line.e