7.02a, 7.02b - Finding equations of lines

7.02a, 7.02b - Finding equations of lines

Can one point tell you the equation of a line? Find the extra information you need, then build straight-line equations from a gradient, two points or a parallel line.

A point and a gradient

Infinitely many lines pass through (2,10)(2,10). If the gradient must also be 33, how can only one line remain?

The gradient fixes the direction. Sliding a line with that direction up or down changes its yy-intercept. Requiring it to pass through a particular point fixes that remaining choice.

You already know y=mx+cy=mx+c. A given point (a,b)(a,b) means x=ax=a and y=by=b must satisfy the equation together. Substitute both coordinates and the given gradient, solve for cc, then write the equation with variable xx and yy.

Worked example 1: finding the missing intercept

Question. Find the equation of the line with gradient 33 through (2,10)(2,10).

Plan. Put the gradient and point into y=mx+cy=mx+c, then solve for the intercept.

y=3x+cinsert the gradient10=3(2)+cthe point lies on the linec=4subtract 6y=3x+4write the general equation\begin{aligned} y&=3x+c &&\text{insert the gradient}\\ 10&=3(2)+c &&\text{the point lies on the line}\\ c&=4 &&\text{subtract }6\\ y&=3x+4 &&\text{write the general equation} \end{aligned}

Answer. y=3x+4y=3x+4.

Check. Substituting x=2x=2 gives 1010, as required. Moving one unit right gives (3,13)(3,13), a rise of 33.

Change only the sign of the point's xx-coordinate. The intercept must change even though the gradient does not.

Worked example 2: a point left of the axis

Question. Find the equation of the line with gradient 33 through (−2,10)(-2,10).

Plan. Substitute the negative input carefully, then isolate cc.

10=3(−2)+csubstitute both coordinates10=−6+cc=16add 6y=3x+16\begin{aligned} 10&=3(-2)+c &&\text{substitute both coordinates}\\ 10&=-6+c\\ c&=16 &&\text{add }6\\ y&=3x+16 \end{aligned}

Answer. y=3x+16y=3x+16.

Check. Starting at (−2,10)(-2,10), moving two units right raises the height by 66. At x=0x=0, the height is therefore 1616, confirming the intercept independently.

Watch out

The point (2,10)(2,10) does not make the intercept 1010: its xx-coordinate is not zero. Use 10=3(2)+c10=3(2)+c, rather than writing y=3x+10y=3x+10.

A non-vertical line is fixed by one point and one finite gradient. A vertical line instead has equation x=ax=a; it cannot be written as y=mx+cy=mx+c.

A line through two points

Two distinct points give the direction as well as the position of a straight line.

Find the gradient from the change in height divided by the change in horizontal position. Keep the subtraction order the same in numerator and denominator. You can go from the first point to the second, or reverse both changes: the quotient is the same. Then use either point to find the intercept.

Gradient between two points

For distinct points (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) with x1≠x2x_1\ne x_2,

m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}

If x1=x2x_1=x_2, the line is vertical: x=x1x=x_1. Not on either 2027 formula sheet: learn the relationship.

Worked example 3: rising between two points

Question. Find the equation of the line through (1,3)(1,3) and (5,11)(5,11).

Plan. Find the gradient, then substitute one point into y=mx+cy=mx+c.

m=11−35−1=2vertical change divided by horizontal change3=2(1)+cuse the first pointc=1y=2x+1\begin{aligned} m&=\frac{11-3}{5-1}=2 &&\text{vertical change divided by horizontal change}\\ 3&=2(1)+c &&\text{use the first point}\\ c&=1\\ y&=2x+1 \end{aligned}

Answer. y=2x+1y=2x+1.

Check. The unused point gives 2(5)+1=112(5)+1=11. Both points satisfy the equation, so the line is fixed correctly.

Keep the first point and both horizontal coordinates, but lower the second point. This changes the sign of the gradient.

Worked example 4: falling between two points

Question. Find the equation of the line through (1,3)(1,3) and (5,−5)(5,-5).

Plan. Use the same coordinate order for both changes, then find cc.

m=−5−35−1=−2height decreases to the right3=−2(1)+cc=5add 2y=−2x+5\begin{aligned} m&=\frac{-5-3}{5-1}=-2 &&\text{height decreases to the right}\\ 3&=-2(1)+c\\ c&=5 &&\text{add }2\\ y&=-2x+5 \end{aligned}

Answer. y=−2x+5y=-2x+5.

Check. The midpoint of the supplied points is (3,−1)(3,-1). Substitution gives −2(3)+5=−1-2(3)+5=-1, so the equation also passes through their midpoint.

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