# Question Video: Finding the Perpendicular Distance between the Given Point and the Straight Line Mathematics

Find the length of the perpendicular drawn from the point π΄ (β1, β7) to the straight line passing through the points π΅ (6, β4) and πΆ (9, β5).

05:10

### 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 πΆ nine, negative five.

Now, this might look quite complicated, but weβre 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 πΆ equals zero is the absolute value of π΄ times π plus π΅ times π plus πΆ all over the square root of π΄ squared plus π΅ squared. Weβre trying to find the length of the perpendicular drawn from the point π΄ with coordinates negative one, negative seven. So, weβll let π be equal to negative one and π be equal to negative seven.

Weβre interested in the straight line that passes through the points π΅ and πΆ. And so, weβre going to need to perform some calculations to do so. We can find the equation of the straight line passing through points π΅ and πΆ 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 ππ₯ plus π. This time π is the slope and π is the π¦-intercept.

In this video, weβre going to use the first formula. It should be quite clear to us that weβre going to need to calculate the slope of the line passing through π΅ and πΆ 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 doesnβt 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 πΆ 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 weβre 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 πΆ 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. Weβre 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 weβre going to subtract two from both sides of our equation so that we have it in the form π΄π₯ plus π΅π¦ plus πΆ equals zero. And so, we find that the equation of our line is a third π₯ plus π¦ plus two equals zero. Weβll now clear some space for the next step.

Comparing the equation of the line π΅πΆ to the equation in our formula, and we find that π΄ is equal to one-third. π΅ is the coefficient of π¦, so thatβs one. And πΆ is the constant, so itβs 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, thatβs a third times negative one, plus π΅ times π, thatβs one times negative seven, plus πΆ, which is two. This is all over the square root of π΄ squared plus π΅ squared. So, thatβs 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. Weβre 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 weβre interested in the absolute value of negative sixteen-thirds, we just get sixteen-thirds. And the denominator is the square root of ten-ninths. Weβll clear some space for the final few steps. Weβll 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, weβre allowed to do this because the square of 10 over the square of 10 is simply one. So, this doesnβt 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 πΆ is eight root 10 over five units.

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