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
