Linear Equations

Lesson

We have now looked at a number of ways of finding the equation of a straight line.

Equation of Lines!

We have:

$y=mx+b$`y`=`m``x`+`b` (gradient-intercept form)

$ay+bx-c=0$`a``y`+`b``x`−`c`=0 (general form)

$y-y_1=m\left(x-x_1\right)$`y`−`y`1=`m`(`x`−`x`1) (point-gradient formula)

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$`y`−`y`1`x`−`x`1=`y`2−`y`1`x`2−`x`1 (two point formula)

It's now time to practice using these different forms.

A line has the equation $3x-y-4=0$3`x`−`y`−4=0.

Express the equation of the line in gradient-intercept form.

What is the gradient of the line?

What is the $y$

`y`-value of the $y$`y`-intercept of the line?

A straight line passes through the point ($0$0, $\frac{3}{4}$34) with gradient $2$2.

Find the equation of the line in the form $y=mx+b$

`y`=`m``x`+`b`.Express this equation in the general form $ax+by+c=0$

`a``x`+`b``y`+`c`=0.Find the $x$

`x`-intercept.

Consider the line with equation: $3x+y+2=0$3`x`+`y`+2=0

Solve for the $x$

`x`-value of the $x$`x`-intercept of the line.Solve for the $y$

`y`-value of the $y$`y`-intercept of the line.Plot the line.

Loading Graph...

Answer the following.

Find the equation, in general form, of the line that passes through $A$

`A`$\left(-12,-2\right)$(−12,−2) and $B$`B`$\left(-10,-7\right)$(−10,−7).Find the $x$

`x`-coordinate of the point of intersection of the line that goes through $A$`A`and $B$`B`, and the line $y=x-2$`y`=`x`−2.Hence find the $y$

`y`-coordinate of the point of intersection.

Relate graphs, tables, and equations to linear, quadratic, and simple exponential relationships found in number and spatial patterns

Investigate relationships between tables, equations and graphs