Equations of Straight Lines

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​−54​x 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.

Question 1

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​.

 

Question 2

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). 

 

Question 3

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.

Question 4

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).

 

Question 5

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​−y1​x2​−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. 

 

Question 6

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). 

 

 

Some Worked Video Examples 

Question 7

Question 8

Question 9