a) Find the gradient of the line x/7+y/11=1.
b) the straight line Y=mx+c is parallel to the line y+ 5/3x=7/3 and passes through the point (2,3). Find the values of m and c
The answer for a is -11/7x
And b us 6 1/3
x/7+y/11=1
multiply each term by 77, the LCD
11x + 7y = 77
7y = -11x + 77
y = (-11/7)x + 11
the gradient or slope is -11/7
(in your answer you included the variable)
b)
y+ 5/3x=7/3 , or
y = (-5/3)x + 7/3
the slope is -5/3 , so
Y=mx+c ----> Y = (-5/3)x + c
if (2,3) lies on it,
3 = (-5/3)(2) + c
c = 3 + 10/3
c = 19/3
so you are correct with 6 1/3
To find the gradient of the line in question a), we need to rearrange the equation into slope-intercept form, which is in the form of "y = mx + c", where m is the gradient of the line.
a) Given equation: x/7 + y/11 = 1
Multiply the equation by the least common denominator, which is 77 to eliminate the fractions:
77 * (x/7) + 77 * (y/11) = 77 * 1
11x + 7y = 77
Rearrange the equation to isolate "y":
7y = -11x + 77
y = (-11/7)x + 11
Comparing this equation to the slope-intercept form, we can see that the gradient (m) is -11/7.
b) To find the values of m and c in the equation Y = mx + c, given that the line is parallel to the equation y + 5/3x = 7/3 and passes through the point (2,3), we can follow these steps:
1. Find the gradient (m) of the given line, y + 5/3x = 7/3.
Rearrange the equation to isolate "y":
y = (-5/3)x + 7/3
Comparing this equation to the slope-intercept form, we can determine that the gradient (m) is -5/3.
2. Since the line we are looking for is parallel to the given line, it will have the same gradient, which is -5/3. Therefore, m = -5/3.
3. We can use the general equation of a line (y = mx + c) and substitute the coordinates (2,3) into the equation to find the value of c.
y = mx + c
3 = (-5/3)(2) + c
3 = -10/3 + c
3 + 10/3 = c
(9 + 10)/3 = c
19/3 = c
Therefore, the value of c is 19/3.
The values of m and c for the line parallel to y + 5/3x = 7/3 and passing through the point (2,3) are m = -5/3 and c = 19/3.