A line L1 passes through point (1,2) and has a gradient of 5.Another line L2 is perpendicular to L1 and meets it at a point where x=4.find the equation of L2.i got L2 gradient as -5 am i right. How do i use x=4 and the equation.please help
No. L2 has slope -1/5
⊥ lines have slopes which are negative reciprocals
The point-slope equation of L1 is
y-2 = 5(x-1)
So, when x=4, y=17
Using the point (4,17) on L2, its equation is
y-17 = -1/5 (x-4)
L1: (1, 2), (4, y), m = 5.
m = (y-2)/(4-1)
5 = (y-2)/3
y-2 = 15
Y = 17.
L2: (4, 17), m = -1/5.
Y = mx+b
17 = (-1/5)4+b
b = 17+4/5 = 17.8
Y = (-1/5)x+17.8
To find the equation of line L2, we first need to determine its gradient. Since line L2 is perpendicular to line L1, it will have a negative reciprocal gradient.
The gradient of L1 is given as 5. The negative reciprocal of 5 is -1/5. Therefore, the gradient of L2 is -1/5.
Next, we will use the given information that L2 meets L1 at a point where x = 4. This means that the x-coordinate of the point of intersection is 4. Let's denote the y-coordinate of the point of intersection as y. So, the coordinates of the point of intersection are (4, y).
Now we can use the point-slope form of a line to find the equation of L2. The point-slope form is given as:
y - y1 = m(x - x1)
Where m is the gradient of the line, and (x1, y1) is a point on the line.
Substituting the values into the point-slope form:
y - y1 = -1/5(x - 4)
Now, we need to find the value of y1. We know that the point (1, 2) lies on line L1, so we can substitute these values into the equation:
2 - y1 = -1/5(1 - 4)
Simplifying this equation will give us the value of y1:
2 - y1 = -1/5(-3)
2 - y1 = 3/5
- y1 = 3/5 - 2
- y1 = -7/5
y1 = 7/5
Now we have the value of y1, we can substitute it back into the equation:
y - 7/5 = -1/5(x - 4)
To simplify further, we can multiply both sides by 5 to eliminate the denominators:
5y - 7 = -x + 4
Rearranging the equation to the standard form, y = mx + c, where m is the gradient and c is the y-intercept:
5y = -x + 11
y = (-1/5)x + 11/5
Therefore, the equation of line L2 is y = (-1/5)x + 11/5.