Video Transcript
What is the equation of the line
passing through π΄ with coordinates negative one, three and the intersection of the
lines three π₯ minus π¦ plus five equals zero and five π₯ plus two π¦ plus three
equals zero?
We can begin by recalling that the
intersection of two lines is the point where they meet or cross. We can also recall that there is a
general equation of a line passing through the intersection point of two lines. If two lines have equations π sub
one π₯ plus π sub one π¦ plus π sub one equals zero and π sub two π₯ plus π sub
two π¦ plus π sub two equals zero, then we can write the equation as π sub one π₯
plus π sub one π¦ plus π sub one plus π times π sub two π₯ plus π sub two π¦
plus π sub two equals zero, for any π in the set of real numbers.
So, letβs take the equation three
π₯ minus π¦ plus five equals zero to have the π sub one, π sub one, and π sub one
values. Meaning that π sub one is three,
π sub one is negative one, and π sub one is five. And in the same way, we can take
the second equation for the π sub two, π sub two, and π sub two values, which
will be equal to five, two, and three, respectively. Then, we substitute these into the
general equation of the line, which gives three π₯ minus π¦ plus five plus π times
five π₯ plus two π¦ plus three equals zero.
At this point, what we have worked
out is the general equation of a line which passes through the intersection of the
two lines three π₯ minus π¦ plus five equals zero and five π₯ plus two π¦ plus three
equals zero. And there would in fact be an
infinite number of lines that pass through their intersection point given by the
infinite number of values we could take for π. However, we need to work out the
equation of one specific line. Itβs the line which passes through
this intersection point and the point π΄ with coordinates negative one, three. And so we can substitute the values
π₯ equals negative one and π¦ equals three into this equation. When we do this and simplify, we
have that negative one plus π times four equals zero. Rearranging, we have four π equals
one and π equals one-quarter.
Now that weβve worked out the value
of π, we can substitute it into this general equation. And this gives three π₯ minus π¦
plus five plus one-quarter times five π₯ plus two π¦ plus three equals zero. To simplify, we distribute the
one-quarter across the parentheses, and then we collect the like terms. This equation of 17 over four π₯
minus two-quarters π¦ plus 23 over four equals zero would be a valid equation of the
line. But it would be a little bit nicer
to write this with integer values for the coefficients of π₯ and π¦. So, by multiplying through by four,
we have 17π₯ minus two π¦ plus 23 equals zero. And thatβs the answer for the
equation of the line which passes through the point π΄ and the intersection of the
two given lines.
Notice that another way in which we
couldβve solved this example without using this general equation form is by first
solving the equations of the two lines simultaneously to find the intersection
point. We would then have needed to
calculate the slope of the line between the two points and used this along with the
pointβslope form of a line to obtain the equation that we found.