### Video Transcript

If the length of the perpendicular drawn from the point negative five, π¦ to the straight line negative 15π₯ plus eight π¦ minus five equals zero is 10 length units, find all the possible values of π¦.

We begin by recalling that we can find the perpendicular distance from a point with Cartesian coordinates π, π to a line in the form π΄π₯ plus π΅π¦ plus πΆ equals zero. Using the formula π is equal to the absolute value of π΄π plus π΅π plus πΆ all over the square root of π΄ squared plus π΅ squared.

Now our line is negative 15π₯ plus eight π¦ minus five. So in this case, we let π΄ be equal to negative 15. π΅ is the coefficient of π¦. So thatβs eight. And then πΆ is our constant. Itβs negative five. We then see that we know the length of the perpendicular drawn from the point negative five, π¦. So we let π be equal to negative five. And we can say π is equal to π¦.

Weβre actually told that the length of this perpendicular is 10. So in this case, we can say that π is equal to 10. Letβs substitute everything we have into our formula for the perpendicular distance. π is equal to 10. And this is equal to the absolute value of π΄ times π β so thatβs negative 15 times negative five β plus π΅ times π β thatβs eight times π¦ β plus πΆ, which is negative five. This is all over the square root of π΄ squared plus π΅ squared. So thatβs the square root of negative 15 squared plus eight squared. Negative 15 times negative five is 75.

Then when we add negative five, thatβs the same as simply subtracting five. And we get 70. Similarly, our denominator becomes the square root of 225 plus 64, which is the square root of 289. But actually, we know that the square root of 289 is 17. So our equation is now 10 is equal to the absolute value of 70 plus eight π¦ all over 17.

Weβll make this a little bit nicer by multiplying both sides by 17. And so we find that 170 is equal to the absolute value of 70 plus eight π¦. Weβre almost ready to solve. We need to appreciate that when solving equations involving absolute values, we need to find both a positive and a negative. In this case, we say that either 70 plus eight π¦ is equal to 170 or 70 plus eight π¦ is equal to negative 170.

Weβll solve this first equation by subtracting 70. And so we see that eight π¦ is equal to 100. Then we divide through by eight to find that π¦ is equal to 100 over eight, which simplifies to 25 over two. And so one possible value for π¦ is 25 over two.

Now weβll solve the second equation in much the same way. We take 70 away from both sides. This time though, subtracting 70 from negative 170 gives us negative 240. We divide through by eight. And we find that π¦ is equal to negative 30. And so our two possible values for π¦ are negative 30 and 25 over two.