Question Video: Finding the Equation of a Straight Line Mathematics

Find the equation of the straight line which passes through the point of intersection of the two lines βˆ’4π‘₯ + 15𝑦 = βˆ’15 and βˆ’4π‘₯ + 3𝑦 = 14 and is parallel to the straight line 𝐫 = <4, 0> + π‘˜<5, βˆ’4>.

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Video Transcript

Find the equation of the straight line which passes through the point of intersection of the two lines negative four π‘₯ plus 15𝑦 equals negative 15 and negative four π‘₯ plus three 𝑦 equals 14 and is parallel to the straight line 𝐫 equals four, zero plus π‘˜ times five, negative four.

The two bits of information we have about the straight line whose equation we want to solve for then is that it passes through the point of intersection of these two lines and that it’s parallel to this line. In general, we can write the equation of a straight line like this. The 𝑦-coordinate of a line is equal to the slope π‘š multiplied by the corresponding π‘₯-coordinate plus the 𝑦-intercept called 𝑏. For a given line then, if we’re able to determine its slope and 𝑦-intercept, then we have defined that line. When we consider the equation of the straight line we want to solve for, we can start by finding out its slope π‘š.

Recall that we’re told that this line is parallel to this given straight line. A line written this way is said to be in vector form. When that’s true, this first vector represents a point on the line and the second vector, the one here by which we’re multiplying π‘˜, gives a vector that’s parallel to the line. That is, a vector where we go five units in the positive π‘₯-direction and four units in the negative 𝑦-direction is parallel to this line. And therefore, it’s parallel to the line whose equation we want to solve for.

In general, the slope of a line equals the amount the line changes in the 𝑦-dimension divided by the corresponding change in the π‘₯-dimension. For our line written in vector form, we see that that ratio is negative four over five. And so we can use this value for π‘š in the equation of our straight line. The last thing we need then to specify our line’s equation is the value of the 𝑦-intercept 𝑏. We can solve for 𝑏 if we know a point that this line passes through. Knowing those coordinates, we can substitute in the π‘₯- and 𝑦-value of that pair and then solve this resulting equation for 𝑏.

Recall we’re told that our straight line passes through the point of intersection of these two lines. If we can find that point then, we’ll have π‘₯- and 𝑦-values we can substitute in here to solve for 𝑏. Looking at the equations of these two lines that intersect, we know that because they intersect, there is some π‘₯- and 𝑦-value, which is true for both equations. This means that we have two independent equations as well as two unknowns to solve for.

We can start solving for these unknowns by subtracting the second equation from the first. When we do this, for the π‘₯-value, we have negative four π‘₯ minus negative four π‘₯, that gives us zero, while for the 𝑦-value, we have 15𝑦 minus three 𝑦. That’s positive 12𝑦. And then, lastly, this equals negative 15 minus 14. That’s negative 29. From this, we see that 𝑦 equals negative 29 over 12. That’s the 𝑦-coordinate of the point of intersection of these two lines. To solve for the corresponding π‘₯-value, we take this 𝑦-value and substitute it in to either of these two equations.

Say that we choose the second equation for this substitution. That gives us negative four π‘₯ plus three times negative 29 over 12 equals 14, or on the left-hand side, negative four π‘₯ minus 29 over four. If we add positive 29 over four to both sides and then divide both sides of that equation by negative four, we get this expression. And we can then note that 14 is the same as 56 divided by four and then that 56 plus 29 equals 85 and then 85 over four divided by negative four equals negative 85 over 16.

This then is the π‘₯-value of the point of intersection of the two lines given to us. And since we’ve already solved for the corresponding 𝑦-value, we can now write out the coordinate pair of that point. Like we mentioned, we’re able to use these π‘₯- and 𝑦-values in our line’s equation to solve for 𝑏. Substituting those values in gives us this expression, and if we then multiply negative four-fifths by negative eighty-five sixteenths, we get a result of 340 over 80. And note that if we divide both the top and the bottom of this fraction by 10, the zeros effectively cancel out.

We can write then that negative 29 over 12 equals 34 over eight plus 𝑏. We can then subtract 34 over eight from both sides. And note that if we multiply top and bottom of this fraction by two and we multiply numerator and denominator of this fraction by three, then both of these fractions will have the same denominator. So we can combine them. Negative 58 minus 102 is negative 160. And then we know that the number eight divides into 160 exactly 20 times, while it divides into 24 exactly three times. Our simplified value for 𝑏 then is negative 20 over three.

Finally then, we can substitute this value for 𝑏 into the equation of our line. And we see that 𝑦 equals negative four-fifths π‘₯ minus 20 over three. This is our answer in slope–intercept form, while an alternative answer form is to bring both variables to the same side of our equation. If we add four-fifths π‘₯ to both sides, we get this equivalent expression.

The equation of the line that satisfies the conditions given in the problem statement is 𝑦 plus four-fifths π‘₯ equals negative 20 divided by three.

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