Exercise 26 Page 326

To determine if the graph of the given equation is a horizontal or vertical line, let's first consider how the given equation would be written in standard form, Ax+ By= C. x=-1.8 ⇔ 1x+ 0y= -1.8Normally, when graphing a line using the standard form, we would substitute 0 for y and solve for x. Then, we would substitute 0 for x and solve for y. Let's try this method using our standard form equation. We will start with the x-intercept.

1x+0y=-1.8

Substitute

y= 0

1x+0( 0)? =-1.8

ZeroPropMult

Zero Property of Multiplication

1x+0? =-1.8

IdPropAdd

Identity Property of Addition

1x? =-1.8

IdPropMult

Identity Property of Multiplication

x=-1.8

Everything is going smoothly so far. The x-intercept lies on the point (-1.8,0). Now let's try the y-intercept.

1x+0y=-1.8

Substitute

x= 0

1( 0)+0y? =-1.8

ZeroPropMult

Zero Property of Multiplication

0+0? =-1.8

AddTerms

Add terms

0≠ -1.8

When we try to find the y-intercept of the given equation, we create a contradiction. This means that the line does not cross the y-axis, and must be vertical. Note that any time we have an equation in the form x=a, where a is a real number, the graph of x=a will be a vertical line.