(a) On the grid above, draw a straight line with a gradient of –3 that passes through the point (–2, 0).

Question

(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.

Answer 1

(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).