4. The straight line, L1, has equation y = x – 2.
(a) Write down the y intercept of L1.
(1)
(b) Write down the gradient of L1.
(1)
The line L2 is perpendicular to L1 and passes through the point (3, 7).
(c) Write down the gradient of the line L2.
(1)
(d) Find the equation of L2. Give your answer in the form ax + by + d = 0 where a, b, d .
(a) The y-intercept of L1 is -2.
(b) The gradient of L1 is 1.
(c) Since L2 is perpendicular to L1, the gradient of L2 is the negative reciprocal of 1, which is -1.
(d) To find the equation of L2, we can use the point-slope form:
y - y1 = m(x - x1)
Where (x1, y1) is the point (3, 7) and m is the gradient of L2, which is -1.
Using the point (3, 7):
y - 7 = -1(x - 3)
Simplifying:
y - 7 = -x + 3
Rearranging the equation:
x + y - 10 = 0
So, the equation of L2 is x + y - 10 = 0.
(a) The y-intercept of line L1 can be found by setting x = 0 in the equation y = x - 2:
y = 0 - 2 = -2.
Therefore, the y-intercept of L1 is -2.
(b) The gradient of line L1 can be found by comparing the equation y = x - 2 to the general equation y = mx + c, where m represents the gradient:
Comparing the two equations we see that m (the gradient) is 1.
Therefore, the gradient of L1 is 1.
(c) Since L2 is perpendicular to L1, the product of their gradients must be -1.
Therefore, the gradient of L2 = -1 / 1 = -1.
(d) We know that the line L2 passes through the point (3, 7) and has a gradient of -1. We can use the point-slope form of a line to find the equation of L2:
y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the gradient.
Using the point (3, 7), we have:
y - 7 = -1(x - 3)
y - 7 = -x + 3
x + y - 10 = 0
Therefore, the equation of L2 in the form ax + by + d = 0 is x + y - 10 = 0.