E2.11, E3.2, E3.5 - Linear equations in different forms

E2.11, E3.2, E3.5 - Linear equations in different forms

Different-looking equations can describe the same straight line. You will rearrange and obtain line equations for E3.5, draw their graphs for E3.2, and recognise and sketch the linear family in E2.11. Choosing a useful form makes gradients, intercepts and special cases easier to see.

Rearranging a linear equation

Do 2x+3y=122x+3y=12 and y=−23x+4y=-\frac23x+4 describe different lines? At first they look different, but subtracting 2x2x and dividing by 33 transforms the first into the second without changing which coordinate pairs satisfy it.

General linear equation

A straight-line equation can be written as

ax+by=c,ax+by=c,

where aa, bb and cc are constants and a,ba,b are not both zero. The terms contain only first powers of xx and yy, with no product xyxy. In this lesson the coefficients are rational numbers.

If b≠0b\ne0, isolate yy by subtracting axax from both sides and dividing every term by bb. This gives y=−abx+cby=-\frac abx+\frac cb. You can then read the gradient and y-intercept. The letters a,b,ca,b,c here name the coefficients in the general form; the constant on the right is not automatically the y-intercept.

Worked example 1: a positive y-coefficient

Question. Rearrange 2x+3y=122x+3y=12 into y=mx+cy=mx+c. State its gradient and y-intercept.

Plan. Leave the yy-term alone, then divide the whole equation by its coefficient.

3y=12−2xsubtract 2xy=4−23xdivide both terms by threey=−23x+4write the gradient term first\begin{aligned} 3y&=12-2x && \text{subtract }2x\\ y&=4-\frac23x && \text{divide both terms by three}\\ y&=-\frac23x+4 && \text{write the gradient term first} \end{aligned}

Answer. y=−23x+4y=-\frac23x+4, with gradient −23-\frac23 and y-intercept 44.

Check. Multiplying the final equation by 33 and adding 2x2x returns 2x+3y=122x+3y=12. At x=0x=0, the original gives 3y=123y=12, confirming the intercept 44.

Change only the sign of the yy-coefficient. Dividing by a negative number changes the signs of both terms on the other side.

Worked example 2: a negative y-coefficient

Question. Rearrange 2x−3y=122x-3y=12 into y=mx+cy=mx+c. State its gradient and y-intercept.

Plan. Isolate −3y-3y, then divide every term by −3-3.

−3y=12−2xsubtract 2xy=−4+23xdivide by negative threey=23x−4collect the required form\begin{aligned} -3y&=12-2x && \text{subtract }2x\\ y&=-4+\frac23x && \text{divide by negative three}\\ y&=\frac23x-4 && \text{collect the required form} \end{aligned}

Answer. y=23x−4y=\frac23x-4, with gradient 23\frac23 and y-intercept −4-4.

Check. Substitute (3,−2)(3,-2): the original gives 6−3(−2)=126-3(-2)=12, and the rearranged equation gives 2−4=−22-4=-2. Reversing the rearrangement also returns the original equation.

When b=0b=0, the condition on the coefficients forces a≠0a\ne0. Then ax=cax=c gives the vertical line x=c/ax=c/a; dividing by bb would be invalid. When a=0a=0 and b≠0b\ne0, the equation gives the horizontal line y=c/by=c/b. For example, 4x=124x=12 means x=3x=3, while 4y=124y=12 means y=3y=3.

If both coefficients were zero, 0x+0y=00x+0y=0 would describe every point in the plane, and 0x+0y=50x+0y=5 would describe no points. Neither is one straight line. This is why the definition excludes that pair of coefficients.

Watch out

From −3y=12−2x-3y=12-2x, the line y=4+23xy=4+\frac23x is wrong: it divides the second term by −3-3 but loses the sign change in the first. Divide every term by the same non-zero number.

A different coefficient gives this fresh check.

Drawing and sketching different forms

The graph of x+2y=6x+2y=6 contains all coordinate pairs satisfying that equation. You can rearrange to y=3−12xy=3-\frac12x and make a table, just as before. Another efficient choice is to find the two axis intercepts: on the xx-axis, y=0y=0; on the yy-axis, x=0x=0.

A drawn graph needs labelled axes, even numbered scales and accurately plotted points joined with a ruler. A sketch shows the line's direction, axis crossings and other important features with labels; its distances need not be to scale. Even a sketch must show a positive crossing on the positive side of the axis.

[DIAGRAM: asset_name: Plotting general line equations; asset_slug: general_line_plotting; caption: Two intercepts fix the left line; the line through the origin needs another distinct point.; description: Two equal-scale numbered grids. The left plots x plus two y equals six through zero comma three, six comma zero and negative two comma four. The right plots x plus two y equals zero through negative four comma two, zero comma zero and four comma negative two. Each line is extended across its grid.]
Two intercepts fix the left line

Worked example 3: using distinct intercepts

Question. Find the axis intercepts of x+2y=6x+2y=6, then draw its graph on suitable axes.

Plan. Set one coordinate to zero at a time, then use a third point as a check.

y=0:x=6find the x-axis crossingx=0:2y=6⇒y=3find the y-axis crossingx=−2:−2+2y=6⇒y=4a third point\begin{aligned} y=0:&\quad x=6 && \text{find the x-axis crossing}\\ x=0:&\quad 2y=6\Rightarrow y=3 && \text{find the y-axis crossing}\\ x=-2:&\quad -2+2y=6\Rightarrow y=4 && \text{a third point} \end{aligned}

Answer. Plot (6,0)(6,0), (0,3)(0,3) and (−2,4)(-2,4) on numbered axes, then rule and extend the line through them, as shown on the left.

Check. From (−2,4)(-2,4) to (0,3)(0,3), a run of 22 gives rise −1-1. From (0,3)(0,3) to (6,0)(6,0), a run of 66 gives rise −3-3. Both give gradient −12-\frac12.

Change only the constant on the right from 66 to 00. Both intercept calculations now return the origin; they have not supplied two distinct points.

Worked example 4: a line through the origin

Question. Find three points on x+2y=0x+2y=0, then draw its graph.

Plan. Use the origin and choose two non-zero inputs, rather than counting the origin twice.

x=0:2y=0⇒y=0the shared axis crossingx=−4:2y=4⇒y=2a point left of the originx=4:2y=−4⇒y=−2a point right of the origin\begin{aligned} x=0:&\quad 2y=0\Rightarrow y=0 && \text{the shared axis crossing}\\ x=-4:&\quad 2y=4\Rightarrow y=2 && \text{a point left of the origin}\\ x=4:&\quad 2y=-4\Rightarrow y=-2 && \text{a point right of the origin} \end{aligned}

Answer. Plot (−4,2)(-4,2), (0,0)(0,0) and (4,−2)(4,-2), then rule and extend the line, as shown on the right. A sketch would label the axes and origin and show the line falling through quadrants II and IV.

Check. Rearranging gives y=−12xy=-\frac12x: the gradient is −12-\frac12 and the y-intercept is zero, agreeing with the drawn line.

Continue this lesson.

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