Linear Equations I

Hong Kong

Stage 1 - Stage 3

Lesson

As we previously saw, a linear equation describes a line which we can draw from a table of values. Each point that we plot onto the $xy$`x``y`-plane represents just one of the infinitely many points that make up the line. Is there a way to know if a given point, say $\left(a,b\right)$(`a`,`b`), lies on the line without having to graph the line?

Checking if a point lies on a line

We can check if a point lies on a line by directly substituting the coordinates of the point into the equation of a line. If both sides are equal, then we claim that the point lies on the line. If both sides are not equal, then we claim that the point does not lie on the line.

Let's look at an example of substituting a point into an equation.

Does $\left(4,-36\right)$(4,−36) lie on the line $y=-8x-4$`y`=−8`x`−4?

**Think:** Checking to see whether a point lies on a line requires us to substitute the point $\left(4,-36\right)$(4,−36) into the equation $y=-8x-4$`y`=−8`x`−4 and seeing if both sides of the equation are equal.

**Do:** Substituting the point $\left(4,-36\right)$(4,−36) means to replace $x$`x` with $4$4 and $y$`y` with $-36$−36 into the equation $y=-8x-4$`y`=−8`x`−4.

Let's first substitute $x=4$`x`=4 into the right hand side and see if it equals our expected value of $y$`y`:

$y$y |
$=$= | $-8x-4$−8x−4 |

$=$= | $-8\times4-4$−8×4−4 | |

$=$= | $-32-4$−32−4 | |

$=$= | $-36$−36 |

Of course, replacing $y$`y` with $-36$−36 simply gives us $-36$−36 and we can see that both sides of the equation are equal at the point $\left(4,-36\right)$(4,−36). So the point lies on the line.

The strategy of substituting the coordinate of a point to determine whether it lies on the line may be used in reverse to find values of an unknown variable. Consider the following example.

The equation of a line is $3x-y-4=0$3`x`−`y`−4=0 and it passes through point $\left(3,k\right)$(3,`k`).

Find the value of $k$`k`.

**Think**: We know that we can substitute the point $\left(3,k\right)$(3,`k`) into the equation because we're told the point lies on the line.

**Do**: To substitute, we replace $x$`x` with $3$3 and $y$`y` with $k$`k` in the equation $3x-y-4=0$3`x`−`y`−4=0.

$3x-y-4$3x−y−4 |
$=$= | $0$0 |

$3\times3-k-4$3×3−k−4 |
$=$= | $0$0 |

$9-k-4$9−k−4 |
$=$= | $0$0 |

$5-k$5−k |
$=$= | $0$0 |

$5$5 | $=$= | $k$k |

Now the last equation is the solution, and we can swap each side to write it as $k=5$`k`=5.

Does $\left(-6,3\right)$(−6,3) lie on the line $y=-\frac{x}{2}$`y`=−`x`2?

No

AYes

B

**Which of the following points lie on the line $y=9+\frac{x}{2}$ y=9+x2?**

**$\left(4,11\right)$(4,11)****A****$\left(2,9\right)$(2,9)****B****$\left(11,4\right)$(11,4)****C****$\left(2,10\right)$(2,10)****D**

The point $\left(-4,k\right)$(−4,`k`) lies on the line $x+2y-4=0$`x`+2`y`−4=0.

Find the value of $k$

`k`.