C3.6; E3.6 - Parallel line equations

C3.6; E3.6 - Parallel line equations

Parallel lines share a direction but occupy different positions. You will recognise equal gradients, find the equation of a parallel line through a stated point and check that it meets both conditions, as required by C3.6 and E3.6.

Parallel lines have equal gradients

Can you change the equation y=2x+1y=2x+1 so that its line never meets the first one? Changing the constant to −3-3 produces y=2x−3y=2x-3. At each input, the second output is four lower. Both lines rise two for every step right, so the gap never closes.

[DIAGRAM: asset_name: Same gradient, different intercepts; asset_slug: parallel_gradients; caption: Equal rises for equal runs keep the lines apart.; description: One numbered Cartesian grid shows the distinct parallel lines y equals two x plus one and y equals two x minus three. Matching triangles on the two lines both have run one and rise two. The y-intercepts one and negative three are highlighted.]
Equal rises for equal runs keep the lines apart.

Parallel lines and gradients

Distinct non-vertical straight lines are parallel exactly when their gradients are equal. In y=mx+cy=mx+c, equal mm and different cc give distinct parallel lines. Equal mm and equal cc give the same line, also called coincident lines.

Why will different intercepts stay separate? For the same input, subtracting the two outputs cancels the identical mxmx terms. Only the fixed difference between their constants remains. It cannot become zero at a special input. The conclusion does not depend on where the diagram stops.

Worked example 1: different intercepts

Question. Are y=2x+1y=2x+1 and y=2x−3y=2x-3 distinct parallel lines?

Plan. Compare gradients, then check that the intercepts differ.

m1=2,m2=2read coefficients of xc1=1,c2=−3different intercepts(2x+1)−(2x−3)=4constant non-zero output gap\begin{aligned} m_1&=2,\quad m_2=2 && \text{read coefficients of }x\\ c_1&=1,\quad c_2=-3 && \text{different intercepts}\\ (2x+1)-(2x-3)&=4 && \text{constant non-zero output gap} \end{aligned}

Answer. Yes. Their gradients are equal and their intercepts are different.

Check. Trying to make their outputs equal gives 2x+1=2x−32x+1=2x-3, hence 1=−31=-3, which is impossible. They have no common point.

Now change only the second intercept from −3-3 to +1+1. Equal gradients alone no longer establish two distinct lines.

Worked example 2: the same line twice

Question. Are y=2x+1y=2x+1 and y=2x+1y=2x+1 distinct parallel lines?

Plan. Compare the complete equations, including their constants.

m1=m2=2same gradientc1=c2=1same intercept(2x+1)−(2x+1)=0identical outputs for every input\begin{aligned} m_1&=m_2=2 && \text{same gradient}\\ c_1&=c_2=1 && \text{same intercept}\\ (2x+1)-(2x+1)&=0 && \text{identical outputs for every input} \end{aligned}

Answer. No. They are coincident: both equations describe the same line.

Check. At x=0x=0, both give y=1y=1; at x=2x=2, both give y=5y=5. Since their expressions are identical, this agreement holds for every input.

Distinct horizontal lines are parallel because each has gradient zero. Distinct vertical lines are also parallel, but their gradients are undefined: compare their equations x=kx=k instead of trying to compare numerical gradients.

Watch out

The gradients of y=2x+1y=2x+1 and y=−2x+1y=-2x+1 are not equal. Opposite signs mean different directions, even though both coefficients have size two. Compare signed gradients.

Finding an intercept through a point

Knowing a parallel line's gradient does not fix its position. A stated point supplies the missing information. Keep the known gradient, write y=mx+cy=mx+c, substitute the point's two coordinates and solve for the unknown constant cc.

You already know

A point (x,y)(x,y) lies on a line when its coordinates satisfy the equation. To solve for an unknown constant, use the same balance operations as in a linear equation.

Substitute both coordinates, including their signs. The point gives a value of yy; it is not automatically the y-intercept, which occurs only when x=0x=0.

Worked example 3: a point to the right

Question. Find the equation of the line parallel to y=3x+4y=3x+4 that passes through (2,1)(2,1).

Plan. Keep gradient three, then use the point to find the intercept.

y=3x+cparallel means equal gradient1=3(2)+csubstitute the pointc=1−6=−5solve for the constant\begin{aligned} y&=3x+c && \text{parallel means equal gradient}\\ 1&=3(2)+c && \text{substitute the point}\\ c&=1-6=-5 && \text{solve for the constant} \end{aligned}

Answer. y=3x−5y=3x-5.

Check. At x=2x=2, the new equation gives y=1y=1, as required. Its output is always nine below the original line, because (3x+4)−(3x−5)=9(3x+4)-(3x-5)=9, so the lines stay distinct and parallel.

Change only the sign of the point's first coordinate. The gradient stays fixed, but the intercept must move.

Continue this lesson.

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