1.03a, 1.03b - Straight lines, distances and gradients | OCR A-Level Mathematics A

Turn a line's direction and position into an equation you can use. You will connect gradients, distances and midpoints, construct parallel and perpendicular lines, and find exactly where lines meet.

Gradient and direction

Can you find the line through $(1,4)$ that meets $3x+2y=7$ at a right angle, without relying on a drawing? Start with what a line does: from $A(-2,1)$ to $B(4,4)$ , moving $6$ units right moves $3$ units up.

You already know

Coordinates give horizontal position first and vertical position second. Subtract coordinates in matching orders, and use Pythagoras for a right-angled triangle.

The rise–run triangle turns direction into a number.

[DIAGRAM: asset_name: Rise and run on a straight line; caption: A rise of 3 over a run of 6 gives gradient one half; asset_slug: h240-gradient-line; description: Equal-scale axes show y equals one half x plus 2 through A minus 2 comma 1 and B 4 comma 4. A dashed right triangle has horizontal run 6 and vertical rise 3. The y-intercept is 2.]

Gradient

The gradient of a non-vertical straight line is its change in $y$ per unit increase in $x$ . A positive gradient rises to the right; a negative gradient falls to the right.

Using two distinct points gives the same ratio wherever they lie on the line.

Gradient between two points

For $A(x_1,y_1)$ , $B(x_2,y_2)$ , with $x_1\ne x_2$ ,

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

Not in the formula sheet: learn it.

Keep both subtraction orders consistent: reversing both changes neither the ratio nor the direction.

Worked example 1: a rising line

Question. Find the gradient through $A(-2,1)$ and $B(4,4)$ .

Plan. Divide the vertical change by the corresponding horizontal change.

\begin{aligned} \Delta y&=4-1=3 && \text{vertical change} \\ \Delta x&=4-(-2)=6 && \text{horizontal change} \\ m&=\frac36=\frac12 && \text{rise divided by run} \end{aligned}

Why this order? Both changes describe the same journey from $A$ to $B$ .

Answer. The gradient is $\frac12$ : each unit right gives half a unit up.

Check. Reversing the journey gives $(-3)/(-6)=\frac12$ .

Now change only the sign of the vertical change: keep the horizontal run but put $B$ below $A$ .

Worked example 2: a falling line

Question. Find the gradient through $A(-2,1)$ and $B(4,-2)$ .

Plan. Use the same ordered differences, retaining the negative rise.

\begin{aligned} \Delta y&=-2-1=-3 && \text{vertical change} \\ \Delta x&=4-(-2)=6 && \text{horizontal change} \\ m&=\frac{-3}{6}=-\frac12 && \text{divide the changes} \end{aligned}

Why keep the sign? It records that $y$ decreases as $x$ increases.

Answer. The gradient is $-\frac12$ .

Check. Moving $6$ units right changes $y$ by $6(-\frac12)=-3$ , reaching $-2$ .

A horizontal line has gradient $0$ . A vertical line, such as $x=3$ , has undefined gradient: its horizontal change is zero, so division is impossible. On equal-scale axes, a line's anticlockwise inclination $\theta$ from the positive $x$ -axis satisfies $m=\tan\theta$ ; this is the same right-triangle ratio, not a formula supplied in the sheet. Thus inclinations $45^\circ$ and $135^\circ$ give gradients $1$ and $-1$ .

Watch out

Wrong: $m=(4-(-2))/(4-1)=2$ in Worked example 1. That is run divided by rise. Right: $m=(4-1)/(4-(-2))=\frac12$ .

Your turn changes the size of the fall as well as the coordinates; keep its sign visible.

Equations and drawing a line

An equation describes every point on a line. In $y=mx+c$ , setting $x=0$ gives $y=c$ , so $c$ is the vertical intercept. The diagram above therefore represents $y=\frac12x+2$ .

Equations of a straight line

Through $(x_1,y_1)$ with finite gradient $m$ ,

y-y_1=m(x-x_1)

Equivalent forms are $y=mx+c$ and $ax+by+c=0$ , where $a,b$ are not both zero. If $b\ne0$ , the latter has gradient $-a/b$ . If $b=0$ , it is a vertical line.

Not in the formula sheet: learn it.

Point–gradient form is the gradient quotient rearranged. It saves you guessing the intercept.

Worked example 3: from a point and gradient

Question. Find the line of gradient $3$ through $(2,-1)$ , in general form.

Plan. Substitute into point–gradient form, then collect all terms on one side.

\begin{aligned} y+1&=3(x-2) && \text{substitute point and gradient} \\ y&=3x-7 && \text{expand and simplify} \\ 3x-y-7&=0 && \text{collect terms} \end{aligned}

Why start with $y+1$ ? Subtracting the point's vertical coordinate means $y-(-1)$ .

Answer. $3x-y-7=0$ .

Check. At $(2,-1)$ , $6+1-7=0$ ; rearranging gives gradient $3$ .

Change only the gradient's sign. The line still passes through the same point, but its intercept changes too.

Worked example 4: a negative gradient

Question. Find the line of gradient $-3$ through $(2,-1)$ , in general form.

Plan. Use the same point–gradient construction with the new gradient.

\begin{aligned} y+1&=-3(x-2) && \text{substitute} \\ y&=-3x+5 && \text{expand} \\ 3x+y-5&=0 && \text{collect terms} \end{aligned}

Why retain the brackets? The gradient multiplies the entire horizontal change.

Answer. $3x+y-5=0$ .

Check. $3(2)+(-1)-5=0$ , and the coefficient of $x$ in $y=-3x+5$ is $-3$ .

If given two points, calculate the gradient first: $(1,2)$ and $(5,10)$ give $m=8/4=2$ , then $y-2=2(x-1)$ , so $y=2x$ . Equal horizontal coordinates instead give a vertical line directly.

To draw $6x+3y-12=0$ , set $y=0$ to get $(2,0)$ and set $x=0$ to get $(0,4)$ . Mark these distinct points on labelled axes and extend the line through them. Alternatively rearrange to $y=-2x+4$ , start at $(0,4)$ and move one unit right, two down. For $x=3$ , mark two points with horizontal coordinate $3$ and join vertically. A graph supplies the reverse information: read a point and calculate rise over run, then construct the equation.

Watch out

Wrong: identify the gradient of $3x+2y=7$ as $3$ . Right: rearrange to $y=-\frac32x+\frac72$ ; the gradient is $-\frac32$ .

Find a new line and use its intercepts to check how you would draw it.

Parallel and perpendicular lines

Direction, rather than position, decides whether lines are parallel or perpendicular. Distinct parallel lines have the same gradient; equal gradients and equal intercepts describe the same line.

Gradient conditions

For non-vertical lines,

\begin{aligned} m_1&=m_2 && \text{parallel directions} \\ m_1m_2&=-1 && \text{perpendicular directions} \end{aligned}

The perpendicular product requires both gradients to be finite. Horizontal and vertical lines are perpendicular to each other.

Not in the formula sheet: learn it.

Why the negative reciprocal? A direction $(u,v)$ rotates through a right angle to $(-v,u)$ . Their gradients are $v/u$ and $-u/v$ , whose product is $-1$ when neither denominator is zero. A useful surprise follows: multiplying both changes by the same positive factor changes the distance travelled but leaves the gradient untouched.

Worked example 5: a parallel line

Question. Find the line through $(1,4)$ parallel to $3x+2y=7$ .

Plan. Read the given gradient, then use it through the new point.

\begin{aligned} y&=-\frac32x+\frac72 && \text{rearrange the given line} \\ y-4&=-\frac32(x-1) && \text{same gradient} \\ 3x+2y-11&=0 && \text{clear fractions and collect} \end{aligned}

Why use the same gradient? Parallel lines make the same vertical change for each horizontal step.

Answer. $3x+2y-11=0$ .

Check. $3(1)+2(4)-11=0$ , and both gradients are $-\frac32$ .

Now change only the requested relationship: perpendicular instead of parallel.

Worked example 6: a perpendicular line

Question. Find the line through $(1,4)$ perpendicular to $3x+2y=7$ .

Plan. Take the negative reciprocal of $-\frac32$ , then use point–gradient form.

\begin{aligned} m&=\frac23 && \text{negative reciprocal} \\ y-4&=\frac23(x-1) && \text{use the point} \\ 2x-3y+10&=0 && \text{clear fractions} \end{aligned}

Why change both sign and reciprocal? Their product must be $-1$ , not $1$ .

Answer. $2x-3y+10=0$ , the line asked for at the start.

Check. $2-12+10=0$ , and $(-\frac32)(\frac23)=-1$ .

For a line perpendicular to $y=4$ , write $x=k$ through the required point; no finite gradient is needed.

Watch out

Wrong: the perpendicular gradient to $2$ is $\frac12$ . Right: it is $-\frac12$ , since $2(-\frac12)=-1$ .

Check the new point and the perpendicular relationship separately.

Midpoints and distances

A midpoint splits each coordinate change in half. A distance combines those changes by Pythagoras, so a negative vertical change does not give a negative length.

Midpoint and distance

For points $A(x_1,y_1)$ and $B(x_2,y_2)$ ,

\begin{aligned} M&=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \\ AB&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \end{aligned}

These work for horizontal and vertical segments too.

Not in the formula sheet: learn it.

Average like coordinates; never average a horizontal coordinate with a vertical one.

Worked example 7: a segment falling to the right

Question. Find the midpoint and length of the segment from $A(-3,2)$ to $B(5,-4)$ .

Plan. Average corresponding coordinates and use the two changes as perpendicular sides.

\begin{aligned} M&=\left(\frac{-3+5}{2},\frac{2-4}{2}\right)=(1,-1) && \text{coordinate averages} \\ AB^2&=8^2+(-6)^2=100 && \text{Pythagoras} \\ AB&=10 && \text{positive square root} \end{aligned}

Why square the signed changes? Pythagoras uses side lengths; squaring removes their directional signs.

Answer. Midpoint $(1,-1)$ , length $10$ .

Check. Both steps $A$ to $M$ and $M$ to $B$ are $(4,-3)$ , each of length $5$ .

Change only the vertical direction: the endpoint is now $(5,8)$ , so the vertical change is $6$ instead of $-6$ .

Worked example 8: same length, different midpoint

Question. Find the midpoint and length from $A(-3,2)$ to $B(5,8)$ .

Plan. Repeat the two coordinate calculations with the new endpoint.

\begin{aligned} M&=\left(\frac{-3+5}{2},\frac{2+8}{2}\right)=(1,5) && \text{coordinate averages} \\ AB^2&=8^2+6^2=100 && \text{Pythagoras} \\ AB&=10 && \text{positive length} \end{aligned}

Why is the length unchanged? Changing the sign of one coordinate change leaves its square unchanged.

Answer. Midpoint $(1,5)$ , length $10$ .

Check. From the midpoint to either endpoint, the squared distance is $4^2+3^2=25$ .

You can reverse an average too. If $M=(2,3)$ and $A=(-1,5)$ , then $(-1+x_B)/2=2$ and $(5+y_B)/2=3$ , giving $B=(5,1)$ .

Watch out

Wrong: $AB=\sqrt{8^2-6^2}$ for a falling segment. Right: add the squared changes, $AB=\sqrt{8^2+(-6)^2}$ .

The next segment uses different changes; check that the midpoint is halfway in both coordinates.

Intersections and exceptional cases

An intersection is one coordinate pair satisfying both equations. Solving simultaneously finds it exactly; a drawing is a check, not a replacement for exact coordinates.

Worked example 9: two sloping lines

Question. Find the intersection of $2x+y=7$ and $x-y=2$ .

Plan. Add to eliminate $y$ , then substitute for its value.

\begin{aligned} 3x&=9 && \text{add the equations} \\ x&=3 && \text{divide by }3 \\ y&=7-2(3)=1 && \text{substitute into the first} \end{aligned}

Why eliminate $y$ ? Its coefficients are opposites already, so no scaling is needed.

Answer. $(3,1)$ .

Check. $2(3)+1=7$ and $3-1=2$ .

Replace only the second line by a vertical one. The intersection can still be found without a gradient for that line.

Worked example 10: one line is vertical

Question. Find the intersection of $2x+y=7$ and $x=4$ .

Plan. Use the given horizontal coordinate directly in the other equation.

\begin{aligned} 2(4)+y&=7 && \text{substitute }x=4 \\ y&=-1 && \text{solve for }y \end{aligned}

Why substitute rather than find two gradients? The vertical line already supplies one coordinate exactly.

Answer. $(4,-1)$ .

Check. $2(4)-1=7$ , and the horizontal coordinate is $4$ .

There are two exceptions to “one intersection”. Combining $y=2x+1$ with $y=2x-3$ gives $1=-3$ : distinct parallel lines, no intersection. Combining $y=2x+1$ with $4x-2y+2=0$ gives $0=0$ : the same line, infinitely many common points. Equal gradients alone do not distinguish these cases.

Exam tip

Give an intersection as a coordinate pair and check it in both equations. An identity means coincident lines; a contradiction means no common point. Neither calls for dividing by zero to produce a coordinate.

All these routes use the same principle: the required line or point must satisfy every stated condition.

Quick Recap

Gradient is vertical change divided by matching horizontal change; vertical lines have undefined gradient.
Use point–gradient form to construct a line and two distinct points to draw it.
Parallel directions have equal gradients; finite perpendicular gradients multiply to $-1$ .
Average coordinates for a midpoint and use Pythagoras for distance.
Solve line equations simultaneously, recognising contradictory and identical equations.

The opening perpendicular line is $2x-3y+10=0$ : it passes through $(1,4)$ and has gradient $\frac23$ , the negative reciprocal of $-\frac32$ .

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1.03a, 1.03b - Straight lines, distances and gradients

Gradient and direction

Equations and drawing a line

Parallel and perpendicular lines

Midpoints and distances

Intersections and exceptional cases

Quick Recap

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