Linear Equations I

Hong Kong

Stage 4 - Stage 5

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 value of 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.

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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.