### Video Transcript

Find the length of the perpendicular drawn from the point π΄ with coordinates negative one, negative seven to the straight line passing through the points π΅ with coordinates six, negative four and πΆ nine, negative five.

Now, this might look quite complicated, but weβre able to find the perpendicular distance from a point to a line. We have a formula. The perpendicular distance from a point with coordinates π, π to a line with the equation π΄π₯ plus π΅π¦ plus πΆ equals zero is the absolute value of π΄ times π plus π΅ times π plus πΆ all over the square root of π΄ squared plus π΅ squared. Weβre trying to find the length of the perpendicular drawn from the point π΄ with coordinates negative one, negative seven. So, weβll let π be equal to negative one and π be equal to negative seven.

Weβre interested in the straight line that passes through the points π΅ and πΆ. And so, weβre going to need to perform some calculations to do so. We can find the equation of the straight line passing through points π΅ and πΆ by using one of the following two formulae. The first is π¦ minus π¦ one equals π times π₯ minus π₯ one, where π₯ one, π¦ one is a point that this line passes through and π is its slope. Similarly, we can use the equation π¦ equals ππ₯ plus π. This time π is the slope and π is the π¦-intercept.

In this video, weβre going to use the first formula. It should be quite clear to us that weβre going to need to calculate the slope of the line passing through π΅ and πΆ before doing anything else. And so, we use the formula π equals π¦ two minus π¦ one over π₯ two minus π₯ one. π₯ one, π¦ one and π₯ two, π¦ two are the coordinates of two points that the line passes through. Now, in reality, it doesnβt much matter which coordinate we choose to be π₯ one, π¦ one and which we choose to be π₯ two, π¦ two. What we absolutely must make sure, though, is that one of the points has coordinates π₯ one, π¦ one not π₯ one, π¦ two for example.

And so, we see the slope of the line passing through points π΅ and πΆ is given by negative five minus negative four over nine minus six. Negative five minus negative four is negative five plus four, which is a negative one. So, the slope of the line weβre interested in is negative one-third. Using point π΅, our formula for the equation of a straight line becomes π¦ minus negative four equals negative one-third times π₯ minus six. Now, we need the equation of our line to be in the form π΄π₯ plus π΅π¦ plus πΆ equals zero.

So, we begin by distributing our parentheses. π¦ minus negative four is π¦ plus four. Then, negative one-third times π₯ minus six is negative one-third π₯ plus two. Weβre going to begin by adding a third of π₯ to both sides of our equation. And we obtain a third π₯ plus π¦ plus four equals two. This is still not quite in the right form, so weβre going to subtract two from both sides of our equation so that we have it in the form π΄π₯ plus π΅π¦ plus πΆ equals zero. And so, we find that the equation of our line is a third π₯ plus π¦ plus two equals zero. Weβll now clear some space for the next step.

Comparing the equation of the line π΅πΆ to the equation in our formula, and we find that π΄ is equal to one-third. π΅ is the coefficient of π¦, so thatβs one. And πΆ is the constant, so itβs two. We already said that π is equal to negative one and π is equal to negative seven. And so, the distance is the absolute value of π΄ times π, thatβs a third times negative one, plus π΅ times π, thatβs one times negative seven, plus πΆ, which is two. This is all over the square root of π΄ squared plus π΅ squared. So, thatβs the square root of a third squared plus one squared.

We simplify a little. And we see that the numerator becomes the absolute value of negative one-third minus seven plus two. And the denominator becomes the square root of a ninth plus one. Weβre going to rewrite negative seven as negative 21 over three and two as six over three. This is so we can create a common denominator and we can add and subtract our fractions where necessary. When we do, we find that negative one-third minus 21 over three plus six-thirds is negative 16 over three.

Similarly, we write one from our denominator as nine over nine. And we find that a ninth plus one is a ninth plus nine over nine, which is 10 over nine. Then, since weβre interested in the absolute value of negative sixteen-thirds, we just get sixteen-thirds. And the denominator is the square root of ten-ninths. Weβll clear some space for the final few steps. Weβll write sixteen-thirds over the square root of ten-ninths as sixteen-thirds divided by ten-ninths. Remember, these are exactly the same.

And then, we recall that we can write the square root of 10 over nine as the square root of 10 over the square root of nine. But, of course, the square root of nine is simply three. Then, we recall that to divide by a fraction, we multiply by its reciprocal. So, this becomes sixteen-thirds times three over root 10. And then, we can cross cancel by three. And we find that the distance is 16 over root 10. We need to rationalise the denominator, so we multiply both the numerator and denominator of our fraction by the square root of 10.

Now, remember, weβre allowed to do this because the square of 10 over the square of 10 is simply one. So, this doesnβt change the size of our fraction. Then, the numerator is 16 root 10. But our denominator is the square root of 10 times the square root of 10, which is 10. Our final step is to simplify by dividing both the numerator and denominator of our fraction by two to get eight root 10 over five. And so, we find the length of the perpendicular drawn from the point π΄ negative one, negative seven to the straight line passing through points π΅ and πΆ is eight root 10 over five units.