The line l has gradient 3 and cuts the x – axis at 4. Find its equation.
Well, if the line cuts the x-axis at 4, we know one point on the line is (4, 0). And since the gradient is 3, we can say the slope is equivalent to going up 3 units for every 1 unit to the right. So, we have a point (4, 0) and a slope of 3. Putting it all together, the equation of the line is "Feeling like a slope-y line today? Try y = 3x - 12!"Keep in mind though, my equations are always clowning around. So, take it with a grain of laughter!
To find the equation of a line, we need to use the slope-intercept form, which is given by:
y = mx + c
Where:
- m represents the gradient (slope) of the line
- c represents the y-intercept (the point where the line crosses the y-axis)
Given that the gradient (m) is 3 and the line cuts the x-axis at 4, we can substitute these values into the equation.
The gradient (m) is 3, so our equation becomes:
y = 3x + c
Since the line cuts the x-axis at 4, it means the point (4, 0) lies on the line. We can substitute these values of x and y into the equation to find the value of c:
0 = 3 * 4 + c
0 = 12 + c
To find the value of c, we can rearrange the equation:
c = -12
Now that we have the value of c, we can substitute it back into the equation:
y = 3x - 12
Therefore, the equation of the line with a gradient of 3 and cuts the x-axis at 4 is y = 3x - 12.
To find the equation of a line with a given gradient and a point on the line, you can use the point-slope form of the equation.
The point-slope form of a linear equation is given by:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line and m is the gradient.
In this case, we are given that the gradient (m) is 3 and the line cuts the x-axis at the point (4, 0).
Let's substitute these values into the point-slope form:
y - 0 = 3(x - 4)
Simplifying, we have:
y = 3x - 12
Therefore, the equation of the line is y = 3x - 12.
If line cuts the x - axis at 4 ,coordinates of this point are ( 4 , 0 )
The "point-slope" form of the equation of a straight line:
y - y1 = m ( x - x1 )
In this case:
x1 = 4 , y1 = 0
gradient m = 3
y - y1 = m ( x - x1 )
y - 0 = 3 ( x - 4 )
y = 3 x - 12