Question Video: Finding the Angle the Perpendicular to a Given Straight Line Makes with the Positive 𝑥-Axis | Nagwa Question Video: Finding the Angle the Perpendicular to a Given Straight Line Makes with the Positive 𝑥-Axis | Nagwa

# Question Video: Finding the Angle the Perpendicular to a Given Straight Line Makes with the Positive 𝑥-Axis Mathematics • Third Year of Preparatory School

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Let 𝑀 be the line on points (0, −8) and (−4, 10) and 𝐿 the perpendicular to 𝑀 that passes through the origin (0, 0). What is the measure of the positive angle that 𝐿 makes with the positive 𝑥-axis? Give your answer to the nearest second.

03:55

### Video Transcript

Let 𝑀 be the line on points zero, negative eight and negative four, 10 and 𝐿 the perpendicular to 𝑀 that passes through the origin zero, zero. What is the measure of the positive angle that 𝐿 makes with the positive 𝑥-axis? Give your answer to the nearest second.

In this question, we are given some information about two lines 𝑀 and 𝐿. We are given two points that 𝑀 passes through. And we are told that 𝐿 is the perpendicular to 𝑀 that passes through the origin. We want to use this information to determine the measure of the positive angle that 𝐿 makes with the positive 𝑥-axis. We need to give our answer to the nearest second.

To answer this question, we can begin by recalling a relationship between the measure of an angle that a line makes with the positive 𝑥-axis and its slope. We can recall that if 𝛼 is the measure of an angle that the line makes with the positive 𝑥-axis, then the tan of 𝛼 must be equal to the slope the of the line. This also holds true for vertical lines if we allow the case where 𝑀 is undefined. We can use this relationship to find the value of 𝛼. It is worth noting that there are infinitely many angles that a line makes with the positive 𝑥-axis, and they are all solutions to this equation. We only want the angle with the smallest positive measure.

Therefore, we want to find the slope of line 𝐿. We can do this by using the fact that it is perpendicular to line 𝑀. We can find the slope of line 𝑀 by using the slope formula: 𝑦 sub two minus 𝑦 sub one over 𝑥 sub two minus 𝑥 sub one, where these are the coordinates of two distinct points on line 𝑀. We can substitute the given coordinates of the two points on line 𝑀 into the formula to obtain 𝑀 equals 10 minus negative eight over negative four minus zero. We can then evaluate this expression to see that the slope of line 𝑀 is negative nine over two. We can find the slope of line 𝐿 by noting it is perpendicular to line 𝑀.

We then recall that we find the slope of a perpendicular line by taking negative the reciprocal of the slope. Therefore, the slope of 𝐿 is two over nine. Another way of thinking about this is to say that the product of the slopes of lines 𝑀 and 𝐿 must be negative one.

We can now substitute the slope of this line into the formula. We do need to be careful to substitute the correct slope of line 𝐿 into this formula, since this is line we want to analyze. Substituting this into the formula gives us that the tan of 𝛼 must be equal to two over nine. Since this is positive, we can solve for 𝛼 by taking the inverse tangent of both sides of the equation. Therefore, 𝛼 is equal to the inverse tan of two over nine, which we can calculate is 12.52 and this expansion continues degrees. At this point, it is always worth double-checking that 𝛼 is positive as required.

We can convert this value into degrees, minutes, and seconds by pressing the conversion button on our calculator. We obtain 12 degrees, 31 minutes, and 43.71 seconds to the nearest hundredth of a second. Rounding this to the nearest second gives us that the line 𝐿 makes an angle of 12 degrees, 31 minutes, and 44 seconds with the positive 𝑥-axis to the nearest second.

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