# 2. The Straight Line

## Slope-Intercept Form of a Straight Line

*m*

The slope-intercept form (otherwise known as "gradient, *y*-intercept" form) of a line is given by:

y=mx+b

This tells us the slope of the line is *m* and the
*y*-intercept of the line is *b*.

### Example 1

The line *y* =
2*x* + 4 has

- slope `m = 2` and
*y*-intercept `b = 4`.

We do not need to set up a table of values to sketch this line. Starting at the *y*-intercept (`y = 4`), we sketch our line by going up `2` units for each `1` unit we go to the right (since the slope is `2` in this example).

To find the *x*-intercept, we let `y = 0`.

2

x+ 4 = 0`x = -2`

We notice that this is a **function**. That is, each value of
*x* that we have gives one corresponding value of
*y*.

See more on Functions and Graphs.

## Point-Slope Form of a Straight Line

*m*

We need other forms of the straight line as well. A useful
form is the **point-slope form** (or point - gradient
form). We use this form when we need to find the equation of a line passing through a point (*x*_{1},* y*_{1}) with slope *m*:

y − y_{1}=m(x−x_{1})

### Example 2

Find the equation
of the line that passes through `(-2, 1)` with slope of `-3`.

Answer

We use:

`y-y_1=m(x-x_1)`

Here,

`x_1= -2``y_1= 1`

`m = -3`

So the required equation is:

`y-1=-3(x-(-2)=-3x-6`

`y=-3x-5`

We have left it in slope-intercept form. We can see the slope
is -3 and the *y*-intercept is -5.

## General Form of a Straight Line

### Need Graph Paper?

Another form of the straight line which we come across is
**general form:**

Ax+By+C= 0

It can be useful for drawing lines by finding the
*y*-intercept (put `x = 0`) and the *x*-intercept
(put `y = 0`).

We also use General Form when finding Perpendicular Distance from a Point to a Line.

### Example 3

Draw the line 2*x* + 3*y* + 12 = 0.

Answer

If `x = 0`, we have: `3y + 12 = 0`, so `y = -4`.

If `y = 0`, we have: `2x + 12 = 0`, so `x = -6`.

So the line is:

Note that the *y*-intercept is `-4` and the *x*-intercept is `-6`.

### Exercises

1. What is the equation of the line perpendicular to the line joining (4, 2) and (3, -5) and passing through (4, 2)?

[Need a reminder? See the section on Slopes of Perpendicular Lines.]

Answer

The line joining `(4, 2)` and `(3, -5)` has slope `m=(-7)/(-1)=7` and is shown as a green dotted line.

We need to find the equation of the magenta (pink) line.

The line **perpendicular** to the green dotted line has slope `-1/7.`

The line through `(4, 2)` with slope `-1/7` has equation:

`y-2=-1/7(x-4)`

`=-x/7+4/7`

`y=-x/7+2 4/7`

2. If `4x − ky = 6` and `6x + 3y + 2 = 0` are perpendicular, what is the value of `k`?

Answer

(2) The slope of 4*x* − *ky* = 6 can be calculated
by re-expressing it in slope-intercept form:

`y=4/kx-6/k`

So we see the slope is `4/k`.

The slope of `6x + 3y + 2 = 0` can also be calculated by re-expressing it in slope-intercept form:

`y=(-6)/3x-2/3=-2x-2/3`

So we see the slope is `-2`.

For the lines to be perpendicular, we need

`4/kxx-2=-1`

This gives `k = 8`.

The resulting line is `4x-8y=6`, which we can simplify to `2x-4y=3.` Here's the graph of the situation:

Perpendicular lines

### Conic section: Straight line

Each of the lines and curves in this chapter are **conic sections**, which means the curves are formed when we slice a cone at a certain angle.

How can we obtain a straight line from slicing a cone?

We start with a **double cone** (2 right circular cones placed apex to apex):

If we slice the double cone by a plane just touching one edge of the double cone, the intersection is a **straight line**, as shown.