Equation of a Line


A linear equation in two variables, often written as \( y = mx + b \), describes a line in the \( xy \)-plane. The line consists of all the solution pairs \( (x_0, y_0) \) that make the equation true.

For example, in the equation, y=2x1 y = 2x -1 , the line would pass through the points (0,1) (0,-1) and (2,3) (2, 3) because these ordered pairs are solutions to the equation, as is every other point on the line.

(1)=2(0)1(3)=2(2)1 \begin{aligned} (-1) = 2(0) -1 \\ (3) = 2(2) -1 \end{aligned}

A line through the points (0,-1) and (2,3) A line through the points (0,-1) and (2,3)

The coefficient m m is the slope of the graph, and the constant b b is the y y -intercept.

The slope of a graph is the ratio between the rate of change of y y and the rate of change of x x , so for two points on a graph p1=(x1,y1) p_1=(x_1,y_1) and p2=(x2,y2) p_2=(x_2,y_2) , slope between them is:

m=y2y1x2x1 m = \frac{y_2 - y_1}{x_2-x_1}


What is the y y -intercept of the line described by 2x3y+9=0 2x - 3y + 9 = 0 ?

If 2x3y+9=0 2x - 3y + 9 = 0 , then 3y=2x+9 3y = 2x + 9 , and so y=23x+93 y = \frac{2}{3}x + \frac{9}{3} , which means the y y -intercept is 3 3 . _\square


What is a a in the point (10,a) (-10, a) , if it falls on the line through the points (2,4) (2,4) and (3,3) (3,3) ?

The slope of the line is m=(3)(4)(3)(2)=1 m = \frac{(3)-(4)}{(3)-(2)}=-1 , so y=x+b y=-x + b for all x x and y y on the graph. This means we can take one of the given points and substitute in its values to find b b .

(3)=(3)+b (3)=-(3) + b

Therefore, b=6 b = 6 , and the equation of the line is y=x+6 y = -x +6 . This means the point (10,a) (-10, a) satisfies the equation a=(10)+6 a = -(-10) +6 . Thus, a=16 a = 16 . _\square

Applications and Extensions

The intersection of the line y=10x+5 y = 10x + 5 with the line through the points (1,9) (1, 9) , (2,20) (2, 20) , is a point (a,b) ( a , b ) . Find the sum of a a and b b .

First, we need to find the equation of the line through (1,9) (1, 9) and (2,20) (2, 20) . Since y2y1x2x1=20921 \frac{y_2 - y_1}{x_2-x_1} = \frac{20-9}{2-1} , the slope of the line must be 11. Further, since 9=11(1)+b 9 = 11(1) + b , the y y -intercept is 2 -2 .

Thus, our two equations are: Line 1: y=10x1+5Line 2: y=11x22 \begin{aligned} \mbox{Line 1: } y &= 10x_1 + 5 \\ \mbox{Line 2: } y &= 11x_2 -2 \end{aligned}

To find the point of intersection (a,b) (a,b) , we solve the system of equations above by equating the two y y 's.

10x+5=11x2 10x + 5 = 11 x -2

Therefore, x=7 x = 7 , and placing 7 7 into either of the equations for xx gives us (7,75) (7, 75 ) as the point of intersection. The answer, then, is 7+75=82 7+75 = 82 . _\square

Note by Arron Kau
7 years, 3 months ago

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I keep wondering at times, why is slope m=y2y1x2x1 m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} and not m=x2x1y2y1 m = \frac{x_{2} - x_{1}}{y_{2} - y_{1}} ? Both of them give same kind of information about the slope, one indicates the change in y per change in x and the other indicates the change in x per change in y. Is it just convention or is there some other reason to it?

Shabarish Ch - 7 years, 3 months ago

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well since m=tanxm=\tan x, its very obvious that m=tanx=y2y1x2x1m=\tan x=\frac{y_2-y_1}{x_2-x_1}. Stay in this pattern is also very good because in higher grades we may need to solve like "What is the equation of the line which is perpendicular with line 5x+2=05x+2=0 and pass through the point (7,0)(7,0)?". Keep it m=tanxm=\tan x rather than cotx\cot x is more simple.

Or another way to explain is as a function f(x)f(x), xx is the one who "change" and yy is the one who "change because xx changed". Then as mm explains the different rate of change of yy against xx, then m=ΔyΔx=y2y1x2x1m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}.

Christopher Boo - 7 years, 3 months ago

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Got it. Thanks.

Shabarish Ch - 7 years, 2 months ago

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What does line1 or line2 represent in three dimensional space?

A Former Brilliant Member - 7 years, 2 months ago

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They represent two lines, since you can write them as: 0z+y=10x1+50z+y=11x22.0z+y=10x_1+5 \\ 0z+y=11x_2-2.

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