Question Video: Finding the Equation of the Given Straight Line | Nagwa Question Video: Finding the Equation of the Given Straight Line | Nagwa

# Question Video: Finding the Equation of the Given Straight Line Mathematics • First Year of Secondary School

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Find the equation of the straight line which passes through the point of intersection of the two lines β13π₯ β 5π¦ = 14 and 2π₯ + 15π¦ = β11 and is parallel to the straight line π₯ + 8π¦ = β14.

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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 13π₯ minus five π¦ equals 14 and two π₯ plus 15π¦ equals negative 11 and is parallel to the straight line π₯ plus eight π¦ equals negative 14.

Okay, so here we have the equation of these two lines. And weβre told that they intersect at some π₯π¦-coordinate point. We want to solve for the equation of the line that passes through that intersection point and is also parallel to this straight line.

In general, one way to write the equation of a straight line is to say that its π¦-coordinate is equal to the slope of the line π multiplied by its π₯-coordinate plus its π¦-intercept π. Since the values of π¦ and π₯ can range all over the coordinate plane, the equation of a given line is specified by defining its slope π and its π¦-intercept π.

In the linear equation we want to find, we can start out by solving for this lineβs slope. Weβll do this by considering the fact that the line whose equation weβre solving for is parallel to this line. Note that if we take this equation and then subtract π₯ from both sides, that gives us this result. And if we then divide both sides by eight, canceling that factor on the left, we find that π¦ equals negative 14 minus π₯ all divided by eight or, written another way, negative π₯ over eight minus seven-fourths.

Weβre interested in this form of our expression because notice that it matches our ππ₯ plus π format. The slope of this line then is the value by which π₯ is being multiplied. Written out completely, this is negative one-eighth. So then, this is the slope of this line. And since the line is parallel to the line whose equation we want to solve for, we know that these two lines have the same slope. The updated form of our lineβs equation then is that π¦ equals negative one-eighth π₯ plus π.

To help us solve for π, weβll want to know just where it is that these two given lines intersect. The coordinates of that point will give us π₯- and π¦-values to substitute into this expression to solve for π. We can look at these two equations as a system of equations. We have two unknowns, π₯ and π¦, and two independent equations. And if we solve for the π₯- and π¦-values that satisfy both of these equations at the same time, then weβll have the coordinates of their point of intersection.

Looking at the values that multiply π₯ and π¦ in both equations, we see that if we multiply this top equation by positive three, then that equation becomes negative 39π₯ minus 15π¦ equals 42. This is useful to us because if we then add these two equations together, we get negative 37π₯ plus zero π¦ equals 31. This implies that π₯ equals negative 31 over 37. And this then is the π₯-value of the point of intersection of our two given lines.

To find the corresponding π¦-value at that point, we can take this π₯-value and substitute it in for π₯ in either one of these equations. Making that substitution into our second equation gives us this result, which we can now solve for π¦. Two times negative 31 over 37 equals negative 62 over 37. And if we then add 62 over 37 to both sides of the equation and divide both sides by 15, we find that π¦ equals negative 11 plus sixty-two thirty-sevenths all divided by 15.

Another way to write negative 11 is as negative 407 divided by 37. We get 407 when we multiply 11 by 37. And this means that when we add this result to 62 over 37, we get negative 345 divided by 37. 15 goes into 345 exactly 23 times. So this fraction overall simplifies to negative 23 over 37. The point at which our two lines intersect then has an π₯-coordinate of negative 31 over 37 and a π¦-coordinate of negative 23 over 37. And because this point lies along the straight line whose equation we want to solve for, we can substitute these π₯- and π¦-values into that lineβs equation to solve for the π¦-intercept π. That gives us this expression. And if we multiply these two fractions together, we get positive 31 over 296. And subtracting this fraction from both sides gives us this result.

Weβd like to add these two fractions together. And we can do so by finding a common denominator. Weβve already seen that 37 when multiplied by eight gives 296. So we can then multiply both numerator and denominator of this first fraction by eight. We have then negative 184 over 296 minus 31 over 296, which comes out to negative 215 over 296. This is our value for π, the π¦-intercept of the equation of the straight line we want to solve for.

So, returning to the equation of that line, if we substitute in this value for π, then we have that π¦ equals negative π₯ over eight minus 215 over 296. This is our answer in slope intercept form. And we can also write this result with the variables π₯ and π¦ on the same side. If we add π₯ plus eight to both sides of this equation, then we have the answer weβll give as the equation of this straight line. π¦ plus π₯ over eight equals negative 215 over 296. Note that this is the same as π¦ plus one over eight times π₯ equals negative 215 over 296.

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