(a) On the grid above, draw a straight line with a gradient of –3 that passes through the point (–2, 0).
(b) Find the equation of this line.
(Total 8 marks)
6. A straight line, L1, has equation x + 4y + 34 = 0.
(a) Find the gradient of L1.
(2)
The equation of line L2is y = mx. L2is perpendicular to L1.
(b) Find the value of m.
(2)
(c) Find the coordinates of the point of intersection of the lines L1 and L2.
(a) To draw a straight line with a gradient of -3 that passes through the point (-2, 0), we can start by plotting the given point on the grid.
Grid:
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Next, we can use the gradient to find another point on the line. The gradient of -3 means that for every 1 unit change in the x-coordinate, there will be a 3 unit change in the y-coordinate.
Starting from the given point (-2, 0), we can move 1 unit to the right and 3 units down to find another point on the line, which will be (1, -3).
Now we can draw a straight line passing through these two points.
Grid:
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The line is represented by the dots (*) on the grid.
(b) To find the equation of the line, we can use the point-slope form of a straight line equation:
y - y1 = m(x - x1)
Using the point (-2, 0) as (x1, y1) and the gradient -3 as m, the equation becomes:
y - 0 = -3(x - (-2))
y = -3(x + 2)
Simplifying the equation, we get:
y = -3x - 6
So the equation of the line is y = -3x - 6.
6. (a) The equation of line L1 is given as x + 4y + 34 = 0.
To find the gradient of L1, we can rearrange the equation to the form y = mx + c, where m is the gradient:
4y = -x - 34
y = -(1/4)x - 34/4
y = -(1/4)x - 8.5
Comparing this equation with y = mx + c, we can see that the gradient of L1 is -1/4.
(b) Line L2 is perpendicular to L1. Perpendicular lines have negative reciprocal gradients.
Since the gradient of L1 is -1/4, the gradient of L2 will be the negative reciprocal, which is 4.
Thus, the value of m for line L2 is 4.
(c) To find the point of intersection between L1 and L2, we can solve the two equations simultaneously:
L1: x + 4y + 34 = 0
L2: y = 4x
Substituting the equation of L2 into L1, we get:
x + 4(4x) + 34 = 0
x + 16x + 34 = 0
17x + 34 = 0
17x = -34
x = -2
Substituting this value of x back into the equation of L2, we get:
y = 4(-2)
y = -8
So the point of intersection of lines L1 and L2 is (-2, -8).
(a) To draw a straight line with a gradient of -3 that passes through the point (-2, 0), we can use the point-slope form of a linear equation:
y - y1 = m(x - x1),
where m is the gradient and (x1, y1) is the given point.
So, substituting the values, we get:
y - 0 = -3(x - (-2))
y = -3(x + 2)
Plotting this on the grid will give us the desired line.
(b) To find the equation of this line, we can rearrange the equation in slope-intercept form (y = mx + c), where m is the gradient and c is the y-intercept.
Expanding the equation from part (a):
y = -3(x + 2)
y = -3x - 6
Therefore, the equation of the line with a gradient of -3 passing through the point (-2, 0) is y = -3x - 6.
6. (a) The equation of line L1 is given as x + 4y + 34 = 0. To find the gradient of L1, we can rearrange the equation in slope-intercept form:
4y = -x - 34
y = -1/4x - 34/4
y = -1/4x - 8.5
Comparing this equation with y = mx + c, we can see that the gradient (m) of L1 is -1/4.
(b) Line L2 is perpendicular to L1, so the product of their gradients should be -1. Let the gradient of L2 be m.
Given the equation of L2 as y = mx, the equation can be rearranged as x = y/m.
To find the value of m, we can substitute x = y/m into the equation of L1:
x + 4y + 34 = 0
(y/m) + 4y + 34 = 0
Multiplying both sides by m to eliminate the fraction:
y + 4my + 34m = 0
Rearranging the equation:
4my + y = -34m
(4m + 1)y = -34m
y = -34m / (4m + 1)
Comparing this equation with y = mx, we can see that m = -34m / (4m + 1).
Simplifying the equation:
(4m + 1)m = -34m
4m^2 + m = -34m
4m^2 + 35m = 0
m(4m + 35) = 0
Either m = 0 or 4m + 35 = 0.
Solving for m, we find:
m = 0 or m = -35/4
Since L2 is a line and not a straight line, the value of m cannot be 0.
Therefore, the value of m is -35/4.
(c) To find the coordinates of the point of intersection of lines L1 and L2, we can equate their equations:
Equation of L1: y = -1/4x - 8.5
Equation of L2: y = (-35/4)x
Substituting y in equation of L2 with y in equation of L1:
-1/4x - 8.5 = (-35/4)x
Simplifying the equation:
-1x - 34 = -35x
34x = 35x - 34
34 = 34x
x = 1
Substituting this value in the equation of L2:
y = (-35/4)(1)
y = -35/4
Hence, the point of intersection of lines L1 and L2 is (1, -35/4).