E4.4 - Different forms of a line (Extended)
Do and describe different lines? E4.4 uses rearrangement to reveal a line's gradient and intercept, then builds an equation from a point and a gradient or from two points. You will give a complete, simplified equation and check the original information.
General form
An equation is a condition on points. Rearranging it with equivalent operations preserves exactly the same solutions, so it preserves the line. In , moving the -term and dividing by 2 reveals the familiar form .
General form of a straight line
The equation , where and are not both zero, describes a straight line. If ,
Its gradient is and its -intercept is . If , the line is vertical: .
The letter in the general form is the right-hand constant; it is not automatically the intercept. The intercept must be read after isolating , or found by setting . These equation forms are not on the supplied formula list for Papers 1–4: learn how to rearrange them.
Worked example 1: a positive coefficient of y
Question. Write in the form , and state its gradient and -intercept.
Plan. Subtract , then divide every term by 2.
Answer. , gradient , -intercept 4. The two opening equations describe the same line.
Check. gives . The point , obtained from the rearranged equation, gives in the original equation too.
Change just the sign of the -coefficient. Dividing by a negative changes the sign of both terms.
Worked example 2: a negative coefficient of y
Question. Write in the form , and state its gradient and -intercept.
Plan. Isolate , then divide the whole right side by .
Answer. , gradient , -intercept .
Check. gives . At , the rearranged equation gives ; in the original.
If and , dividing gives the horizontal line . If , do not divide by : for example, is simply , with undefined gradient.
Watch out
From , dividing only the -term gives a wrong intercept. Both terms must be divided by . The coefficient of in general form is not itself the gradient.
Rearrange a fresh equation before reading off its properties.
Point and gradient
A gradient tells you how a line changes, but many lines can have that gradient at different heights. A point fixes its position. Put the known gradient into , then substitute the point's coordinates to determine .
Why does this work? Every point on the line satisfies the equation. A known point therefore gives an equation with only the intercept still unknown. Solving that small equation locates the one line with the required gradient through the point.
Worked example 3: finding the intercept from a point
Question. Find the equation of the line with gradient 3 through .
Plan. Use , then substitute , .
Answer. .
Check. The coefficient gives gradient 3. Substitution of the original point gives , as required.
Only the point's horizontal coordinate changes sign in the next example. Retain its minus sign when substituting; the intercept will change.
Continue this lesson.
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