Question Video: Finding the Equation of a Straight Line given the Points It Passes Through Mathematics

Given 𝐴(βˆ’3, βˆ’2), 𝐡(0, 5), and 𝐢(2, βˆ’6), find the equation of the straight line that passes through the vertex 𝐴 and bisects the line segment 𝐡𝐢.

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Video Transcript

Given 𝐴: negative three, negative two; 𝐡: zero, five; and 𝐢: two, negative six, find the equation of the straight line that passes through the vertex 𝐴 and bisects the line segment 𝐡𝐢.

So, firstly, we’ve been given the coordinates of one point that this line passes through: the point negative three, negative two. We also know that it bisects the line segment 𝐡𝐢, which means we can work out the coordinates of a second point on the line. If the line bisects the line segment 𝐡𝐢, then it passes through its midpoint. And we know that the midpoint of the line segment of the points π‘₯ one, 𝑦 one and π‘₯ two, 𝑦 two is π‘₯ one plus π‘₯ two over two, 𝑦 one plus 𝑦 two over two. Essentially, the π‘₯-coordinate of the midpoint is the mean of the two π‘₯-coordinates, and the 𝑦-coordinate of the midpoint is the mean of the two 𝑦-coordinates.

The midpoint of 𝐡𝐢 then is zero plus two over two for the π‘₯-coordinate and five plus negative six over two for the 𝑦-coordinate, which simplifies to one, negative one-half. We now know two points that lie on the line we’re looking for, and so we can use the coordinates of these two points to calculate the slope of our line. The slope of the line connecting the points π‘₯ one, 𝑦 one and π‘₯ two, 𝑦 two can be calculated as 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one. It’s change in 𝑦 over change in π‘₯. Substituting the values for our two points, we have negative a half minus negative two for the change in 𝑦 in the numerator and then one minus negative three for the change in π‘₯ in the denominator.

In the numerator, negative a half minus negative two is negative a half plus two, which is one and a half. And in the denominator, one minus negative three. That’s one plus three, which is four. Now this looks very untidy because we have a mixed number in the numerator of our fraction. So we can convert that mixed number, one and a half, to a top-heavy fraction of three over two and think of dividing by four as multiplying by one-quarter. So this simplifies to three over two multiplied by one over four. And then multiplying the numerators and multiplying the dominators gives three-eighths, so the slope of the line is three-eighths.

We now know the slope of our line and the coordinates of two points it passes through, so we can use the point–slope form of the equation of a straight line to calculate its equation: 𝑦 minus 𝑦 one equals π‘šπ‘₯ minus π‘₯ one. It doesn’t matter which of our two points we choose to be π‘₯ one, 𝑦 one. I’m going to use the point 𝐴 negative three, negative two because both of its coordinates are integer values. So we substitute negative two for 𝑦 one, three-eighths for π‘š, and negative three for π‘₯ one to give 𝑦 minus negative two is equal to three-eighths π‘₯ minus negative three. That is, of course, 𝑦 plus two is equal to three-eighths of π‘₯ plus three.

Next, we want to deal with this fraction, and as we have an eight in the denominator, if we multiply the entire equation by eight, we can eliminate this. Doing so gives eight 𝑦 plus 16 on the left-hand side. And on the right-hand side, we’re left with three multiplied by π‘₯ plus three. The next step is going to be to distribute the parentheses on the right-hand side, so we now have eight 𝑦 plus 16 is equal to three π‘₯ plus nine. Finally, we’re going to group all of the terms on the same side of the equation. By subtracting both three π‘₯ and nine from each side of the equation, we can group all the terms on the left-hand side. And this gives our answer to the problem.

For the three given points, the equation of the straight line that passes through point 𝐴 and bisects the line segment 𝐡𝐢 is eight 𝑦 minus three π‘₯ plus seven is equal to zero.

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