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

Find the length of the perpendicular drawn from the origin to the straight line βˆ’3π‘₯ + 4𝑦 βˆ’ 21 = 0 rounded to the nearest hundredth.

04:20

Video Transcript

Find the length of the perpendicular drawn from the origin to the straight line negative three π‘₯ plus four 𝑦 minus 21 equals zero rounded to the nearest hundredth.

So within this question, we’ve been given a point, the origin, and we’ve been given the equation of a straight line: negative three π‘₯ plus four 𝑦 minus 21 equals zero. We’re asked to find the length of the perpendicular between these two. So that is the length of the perpendicular line that starts at the origin and is drawn to meet the straight line.

So in order to answer this question, we need to recall that in fact we have a standard formula for answering questions like this. If we want to calculate the perpendicular distance between a point with coordinates π‘₯ one, 𝑦 one and a line with equation π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐 is equal to zero, then we have the formula 𝑙, which represents the length, is equal to the modulus or absolute value of π‘Žπ‘₯ one plus 𝑏𝑦 one plus 𝑐 all divided by the square root of π‘Ž squared plus 𝑏 squared.

Those vertical lines in the numerator of the formula represent the modulus or absolute value, which means the size of a number ignoring its sign. So essentially, this means its distance from zero. The modulus of seven is seven, but the modulus of negative seven is also seven. So what we need to do for this question is determine the values of π‘Ž, 𝑏, 𝑐, π‘₯ one, and 𝑦 one and then substitute them into the standard formula.

π‘₯ one, 𝑦 one first of all represents the coordinates of the point that we’re drawing this perpendicular from. And in our question, this point is the origin. The origin has the coordinates zero, zero. And therefore, the values of π‘₯ one and 𝑦 one are both zero. In order to find the values of π‘Ž, 𝑏, and 𝑐, we need to look at the equation of the line: negative three π‘₯ plus four 𝑦 minus 21 is equal to zero.

So comparing these two forms, we see that π‘Ž is equal to negative three, 𝑏 is equal to four, and 𝑐 is equal to negative 21. Now we have all the information we need in order to be able to use the formula. So we need to substitute each of the values in the correct places.

So in the numerator first of all, we have 𝑙 is equal to the modulus of π‘Ž multiplied by π‘₯ one, that’s negative three multiplied by zero, plus 𝑏 multiplied by 𝑦 one, so that’s four multiplied by zero, and then plus 𝑐, so that’s plus negative 21. Then we divide by the square root of π‘Ž squared plus 𝑏 squared, so that’s the square root of negative three squared plus four squared. So this simplifies to the modulus of negative 21 over the square root of nine plus 16.

Now, the modulus of negative 21, remember that’s the size of the number ignoring its sign. So the modulus of negative 21 is just 21. In the denominator, nine plus 16 is 25. So we have 21 over the square root of 25. 25 is a squared number, and it has an exact square root, which is five. So our answer simplifies to 21 over five. As a decimal, 21 over five is equal to 4.2.

Now the question has actually asked us to give our answer to the nearest hundredth. And our answer is currently to the nearest tenth, so we just need to add an extra zero in order to give our answer in the requested format. So we have the length of the perpendicular between the origin and the specified line is 4.20 length units. We use length units here because we don’t know the scale of the coordinate grid, so we can’t say that it’s centimetres or millimetres or so on. We have to use the general length units.

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