Video: Finding the Distances between Points and Straight Lines in Two Dimensions

Find the length of the perpendicular drawn from the point 𝐴(1, 9) to the straight line βˆ’5π‘₯ + 12𝑦 + 13 = 0.

02:33

Video Transcript

Find the length of the perpendicular drawn from the point 𝐴 one, nine to the straight line negative five π‘₯ plus 12𝑦 plus 13 equals zero.

So we’re going to answer this question using the formula for calculating the distance between a point and a straight line. So the formula is this. If I have the straight line with equation π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐 is equal to zero and I have a point with coordinates π‘₯ one, 𝑦 one. Then the perpendicular distance between them, 𝑙, is given by the modulus of π‘Žπ‘₯ one plus 𝑏𝑦 one plus 𝑐, all divided by the square root of π‘Ž squared plus 𝑏 squared. So what I need to do is determine the values of π‘Ž, 𝑏, 𝑐, π‘₯ one, and 𝑦 one and then substitute them into the formula.

Let’s look at the straight line first of all. I’m comparing it with π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐 is equal to zero. This shows me that π‘Ž is equal to negative five, 𝑏 is equal to 12, and 𝑐 is equal to 13. Now let’s look at the point 𝐴, which has coordinates one, nine. This tells me that π‘₯ one is equal to one and 𝑦 one is equal to nine. So now I have all the values I need. And it’s just a case of substituting them into this formula for the distance 𝑙.

So we have that 𝑙 is equal to negative five times one plus 12 times nine plus 13, the modulus of that quantity. Then we’re going to divide it by the square root of negative five squared plus 12 squared. This gives us the modulus of negative five plus 108 plus 13 all divided by the square root of 25 plus 144. This gives the modulus of 116 over the square root of 169. Now as 116 is positive, then its modulus is just its own value. So the numerator will be 116. And in the denominator, the square root of 169 is 13 exactly.

So we have our answer to the problem. The length of the perpendicular between the point one, nine and the straight line negative five π‘₯ plus 12𝑦 plus 13 equals 0 is 116 over 13.

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