# ax + by + c = 0

**What are straight line graphs?**

A straight line is represented by an equation of the form $$Ax+By+C=0$$. In this case $$A$$, $$B$$ and $$C$$ are arbitrary constants. $$A$$ and $$B$$ cannot both be $$0$$ and $$x$$ and $$y$$ are variables. Divide the equation by $$B$$ and rearranging the terms yield,

$$y=\frac{-A}{B}x+\frac{-C}{B}$$

Substitute $$\frac{-A}{B}=m$$ and $$\frac{-C}{B}=c$$ to get $$y=mx+c$$

This is slope-intercept form of the equation of the line where $$m$$ is slope of the line and $$y$$-intercept is equal to $$c$$. So, we can say that the equation $$Ax+By+C=0$$ denotes a line with a slope equal to $$ \frac{-A}{B}$$ and $$y$$-intercept equal to $$\frac{-C}{B}$$

For example, $$7x+5y=13$$ and $$-4x+8y=12$$ are linear equations in two variables. The two equations denote the two lines. The slope of the first line is equal to $$\frac{-7}{5}$$ and slope of the second line is $$\frac{1}{2}$$.

**E3.4: Interpret and obtain the equation of a straight-line graph**

The slope-intercept form of a straight-line graph is $$y=mx+c$$, where $$x$$ and $$y$$ are variables, $$m$$ is the gradient or slope and $$c$$ is the $$y$$-intercept.

Straight line graph can be used to describe the behavior of many things in real life. For example, while purchasing vegetables from a store, the price is usually stated in dollars per kilograms. In mathematical terms, you could say that the price is $$y$$ and the number of kilos purchased is $$x$$. Then you could write a mathematical relationship that relates the cost to how much kilos you have purchased. Assume you were purchasing Apples at $$\$1.5 / \text{kilogram}$$.

Rate of apples is $$\$1.5 / \text{kilogram}$$

So, $$y=1.5x$$

We can easily draw the equation of a line on the graph.

**Worked examples of straight line problems**

**Example 1: **Rewrite the equation $$2x+3y=6$$ in the form of $$y=mx+c$$. Find the value of the gradient and the $$y$$-intercept.

**Step 1: Change the given equation in the form of $$y=mx+c$$.**

$$2x+3y=6$$

$$3y=6-2x$$

$$y=\frac{-2}{3}x+2$$

**Step 2: Compare the equation with the general form and find the gradient. **

Compare $$y=mx+c$$ and $$y=\frac{-2}{3}x+2$$

So, we get $$m=\frac{-2}{3}$$

**Step 3: Compare the equation with the general form and find the $$y$$-intercept.**

The equation is $$y=\frac{-2}{3}m+2$$

So, the $$y$$-intercept is $$2$$

Thus, the slope of line is $$\frac{-2}{3}$$ and $$y$$-intercept is $$2$$.

**Example 2: **Find the equation of a straight line that passes through the point $$( 2,5)$$ with gradient $$-2$$. Convert the equation of line to $$ax+by+c=0$$ form.

**Step 1: Recall the equation of the line formed by using a point and a gradient.**

$$y-y_{1}=m\left ( x-x_{1} \right )$$

**Step 2: Substitute the value of points and slope in the general equation of the line.**

$$y-2= -2(x-5)$$

**Step 3: Solve further to get the value.**

$$y-2=-2x+10$$

**Step 4: Add $$2$$ on both the sides of the equation.**

$$y=-2x+12$$

**Step 5: Write the straight-line equation in the required form.**

$$y=-2x+12$$

$$2x+y-12=0$$

Hence, the required equation of line is $$2x+y-12=0$$.

**Example 3: **Find the equation of a straight line in the form of $$ax+by+c=0$$ that passes through the point $$(3,7)$$ with gradient $$-4$$.

**Step 1: Recall the equation of line formed by using a point and a gradient.**

$$y-y_{1}=m(x-x_{1})$$

**Step 2: Substitute the value of points and slope in the general equation of the line.**

$$y-7 = -4x +12$$

**Step 3: Solve further to get the value.**

$$y-7=-4x+12$$

**Step 4: Add $$7$$ on both the sides of the equation.**

$$y=-4x+19$$

**Step 5: Find the straight-line equation in the required form.**

$$y=-4x+19$$

$$4x+y-19=0$$

Hence, the required equation of the line is $$4x+y-19=0$$.