We have now looked at a number of ways of finding the equation of a straight line.
We have:
$y=mx+b$y=mx+b (gradient-intercept form)
$ay+bx-c=0$ay+bx−c=0 (general form)
$y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1) (point-gradient formula)
$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$y−y1x−x1=y2−y1x2−x1 (two point formula)
It's now time to practice using these different forms.
A line has the equation $3x-y-4=0$3x−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=mx+b.
Express this equation in the general form $ax+by+c=0$ax+by+c=0.
Find the value of the $x$x-intercept.
Consider the line with equation: $3x+y+2=0$3x+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.
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.