The line with equation $y=mx+b$y=mx+b has a slope $m$m and a y intercept $b$b . It is important to observe that this form of the line shows $y$y explicitly as a function of $x$x with $m$m and $b$b as constants, different values of $x$x will determine different values of $y$y .
For example, the line, say $L_1$L1, given by $y=3x+3$y=3x+3 has a slope of $3$3 and a $y$y intercept of $3$3. The $y$y intercept can be determined by noting that at $x=0$x=0, $y=3$y=3.
The line $L_2$L2 given in general form as $2x+y-8=0$2x+y−8=0 can be rearranged to $y=-2x+8$y=−2x+8 and the slope $-2$−2 and y intercept $8$8 can be easily determined.
The line $L_3$L3 given by $5x+4y-29=0$5x+4y−29=0 can be rearranged to $4y=29-5x$4y=29−5x and then to $y=\frac{29}{4}-\frac{5}{4}x$y=294−54x with slope $m=-\frac{5}{4}$m=−54 and $y$y intercept $b=7.25$b=7.25.
We will now go through some of the skills in finding lines, intersections and midpoints by considering a number of questions relating to the lines $L_1,L_2$L1,L2 and $L_3$L3. As we answer the questions, check the sketch below to confirm your understanding of each answer.
Is the point $A\left(1,6\right)$A(1,6) on $L_1$L1?
By substituting $\left(1,6\right)$(1,6) into $y=3x+3$y=3x+3 we see that $6=3\times1+3$6=3×1+3. This is true and so the given point is on $L_1$L1.
Find the mid-point $M$M of the line segment between $A$A and $B\left(9,-4\right)$B(9,−4)
The mid-point is simply the point with coordinates as averages of the coordinates of the two given points. So $M=\left(\frac{1+9}{2},\frac{6+-4}{2}\right)$M=(1+92,6+−42) or simplified $M=\left(5,1\right)$M=(5,1).
Show that $L_3$L3 contains the midpoint $M$M
Since $M=\left(5,1\right)$M=(5,1), substitute into $L_3$L3 so that $5\times5+4\times1-29=0$5×5+4×1−29=0, and because the point satisfies $L_3$L3 we know the point is on the line.
Find $P$P, the $x$x intercept of $L_2$L2.
Since $L_2$L2 is given by $2x+y-8=0$2x+y−8=0, the $x$x intercept is found by putting $y=0$y=0. Then $2x-8=0$2x−8=0 and solving for $x$x, we see that $x=4$x=4. The point of intercept is thus $P\left(4,0\right)$P(4,0).
Find the equation of the line $L_4$L4, which passes through $P$P and $M$M.
With $P\left(4,0\right)$P(4,0) and $M\left(5,1\right)$M(5,1), we have two methods to find $L_4$L4. Both methods require finding the slope of the line given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{1-0}{5-4}=1$m=y2−y1x2−x1=1−05−4=1.
Then method 1 makes use of the point slope form of the line. Specifically we know that the equation we are looking for must have the form $y=1x+b$y=1x+b. Since $M\left(5,1\right)$M(5,1) is on this line, it must satisfy it. Thus we can write $1=1\times5+b$1=1×5+b and so with a little thought, $b$b must be $-4$−4. the equation of $L_4$L4 must be $y=x-4$y=x−4.
The second method makes use of the point slope formula $y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1). We know that the slope $m=1$m=1 and choosing one of the known points on the line, say $M\left(5,1\right)$M(5,1), we can determine the equation of $L_4$L4 as $y-5=1\left(x-4\right)$y−5=1(x−4), and this simplifies once again to $y=x-4$y=x−4.
Find the point $Q$Q where $L_1$L1 intersects with $L_4$L4
We can set the two equations up as follows;
$L_1$L1 $y=3x+3$y=3x+3
$L_4$L4 $y=x-4$y=x−4
At the point of intersection the y values of both lines are equal, and this means that, at that moment, the right hand sides of $L_3$L3 and $L_4$L4 are also equal. Thus we can set $3x+3=x-4$3x+3=x−4 and solve to find $x=-3.5$x=−3.5. Putting this value into equation $L_4$L4 we see that $y=-3.5-4=-7.5$y=−3.5−4=−7.5. Thus the coordinates of $Q$Q are $\left(-3.5,-7.5\right)$(−3.5,−7.5).
A line passes through the point $A$A$\left(-4,3\right)$(−4,3) and has a slope of $-9$−9. Using the point-slope formula, express the equation of the line in slope intercept form.
A line passes through the points $\left(4,-6\right)$(4,−6) and $\left(6,-9\right)$(6,−9).
Find the slope of the line.
Find the equation of the line by substituting the slope and one point into $y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1).
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.