Equation of a Straight Line

Use y=mx+c.

1. A straight line has equation y=mx+c, where m is the gradient and c is the y-intercept (the line crosses the y-axis at (0,c)).
2. Here m=3 and c=-2, so y=3x-2.

A point with x-coordinate 0 lies on the y-axis.

1. (0,5) is on the y-axis, so the y-intercept is c=5.
2. With gradient m=-2, y=mx+c gives y=-2x+5.

Substitute the x-coordinate of each point and see whether you get its y-coordinate.

1. When x=1, y=2\times1-3=-1, so (1,-1) lies on the line.
2. The others do not: (1,1) needs y=-1 at x=1, (-1,-1) gives y=-5 at x=-1, and (2,3) gives y=1 at x=2.

Look at which coordinate is the same for both points.

1. Both points have y-coordinate 4, so the line is horizontal.
2. Every point on it has y=4, so its equation is y=4.

Read where the line crosses the y-axis, then use two points where the line passes through grid corners to find the gradient.

1. The line crosses the y-axis at (0,1), so c=1.
2. It also passes through (2,2). Gradient =\frac{\text{change in }y}{\text{change in }x}=\frac{2-1}{2-0}=\frac12.
3. So the equation is y=\frac12x+1.

Write y=-3x+c and use the point to find c.

1. Use y=mx+c with m=-3: y=-3x+c.
2. (2,7) is on the line, so substitute x=2 and y=7: 7=-6+c.
3. So c=13 and the equation is y=-3x+13.

Gradient =\frac{\text{change in }y}{\text{change in }x}.

1. Gradient =\frac{\text{change in }y}{\text{change in }x}=\frac{11-5}{4-1}=\frac63=2.
2. Substitute (1,5) into y=2x+c: 5=2+c, so c=3.
3. The equation is y=2x+3.

Keep the signs when calculating both coordinate changes.

1. Gradient =\frac{-4-5}{4-(-2)}=\frac{-9}{6}=-\frac32.
2. Substitute (-2,5) into y=-\frac32x+c. Since -\frac32\times(-2)=3, this gives 5=3+c, so c=2.
3. The equation is y=-\frac32x+2.

Look at which coordinate is the same for both points.

1. Both points have x-coordinate -3, so the line is vertical.
2. Every point on it has x=-3, so its equation is x=-3.

Rearrange the equation into the form y=mx+c.

Part (a)

1. Make y the subject. Subtract 6x: 3y=15-6x.
2. Divide by 3: y=5-2x, which is y=-2x+5.
3. In y=mx+c the gradient is m, so the gradient is -2.

Part (b)

1. On the y-axis x=0, so 3y=15 and y=5.
2. The line crosses the y-axis at (0,5).

Read where the line crosses each axis, then find the gradient.

1. The line crosses the y-axis at (0,3), so c=3.
2. It crosses the x-axis at (2,0). Gradient =\frac{0-3}{2-0}=-\frac32: as x goes up by 2, y goes down by 3.
3. The equation is y=-\frac32x+3.

Find the equation of the line through the first two points, then use x=5.

1. Gradient =\frac{12-4}{3-(-1)}=\frac84=2.
2. Substitute (-1,4) into y=2x+c: 4=-2+c, so c=6 and the line is y=2x+6.
3. (5,k) is on the line, so k=2\times5+6=16.

Treat the two intercepts as two points on the line.

1. Gradient =\frac{0-(-6)}{4-0}=\frac64=\frac32.
2. The line crosses the y-axis at (0,-6), so c=-6.
3. The equation is y=\frac32x-6.

The change in cost divided by the change in number gives the charge per poster.

1. The 3 extra posters cost £29-14=15 more, so each poster costs £5.
2. 2 posters cost £10 of the £14, so the setup fee is £4.
3. Cost = setup fee + £5 per poster: C=5n+4.

Find the midpoint by averaging each coordinate before using the given gradient.

1. The midpoint is the mean of the coordinates: \left(\frac{-2+6}{2},\frac{1+9}{2}\right)=(2,5).
2. Use y=\frac12x+c and substitute (2,5): 5=1+c, so c=4.
3. The line is y=\frac12x+4.

Split the journey from A to B into 3+1=4 equal steps.

Part (a)

1. AC:CB=3:1, so C is \frac{3}{3+1}=\frac34 of the way from A to B.
2. From A to B: x changes by 6-(-6)=12 and y changes by -3-5=-8.
3. \frac34 of these changes: x by 9 and y by -6.
4. So C=(-6+9,\ 5-6)=(3,-1).

Part (b)

1. Gradient of AB=\frac{-3-5}{6-(-6)}=\frac{-8}{12}=-\frac23.
2. Substitute A(-6,5) into y=-\frac23x+c. Since -\frac23\times(-6)=4, this gives 5=4+c, so c=1.
3. The equation of AB is y=-\frac23x+1.

The point provides a second linear equation in the two constants.

1. (2,11) is on the line, so substituting into y=mx+c gives 11=2m+c.
2. Subtract m+c=7 from 2m+c=11: m=4.
3. Then c=7-4=3, so the line is y=4x+3.

Use the triangle area to find the missing intercept before finding the gradient.

1. Let B=(0,b). Triangle OAB has a right angle at O, base OA=4 and height OB=b.
2. Area =\frac12\times4\times b=2b. So 2b=12, b=6 and B=(0,6).
3. Gradient of AB=\frac{0-6}{4-0}=-\frac32, and AB crosses the y-axis at (0,6).
4. The equation of AB is y=-\frac32x+6.

Write the gradient of AB using a, then set it equal to -2.

1. Gradient =\frac{\text{change in }y}{\text{change in }x}=\frac{-5-3}{4-a}=\frac{-8}{4-a}.
2. Set this equal to -2 and multiply by 4-a: -8=-2(4-a).
3. Expand: -8=-8+2a, so 2a=0 and a=0.

Find the equation of L, then solve it simultaneously with y=-x+7.

Part (a)

1. Gradient =\frac{8-(-1)}{4-(-2)}=\frac96=\frac32.
2. Substitute (-2,-1) into y=\frac32x+c: -1=-3+c, so c=2.
3. The equation of L is y=\frac32x+2.

Part (b)

1. Where the lines meet, both equations are true, so \frac32x+2=-x+7.
2. Add x and subtract 2: \frac52x=5, so x=2.
3. Substitute into y=-x+7: y=5. The lines meet at (2,5).