Straight Lines
Geometry turns into algebra — every line becomes an equation, and slope, distance and angle are read straight off the coefficients
- The meaning of slope as \(m=\tan\theta\), and how inclination and slope are linked.
- Every form of a line's equation — point-slope, two-point, slope-intercept, intercept and normal — and when each is fastest.
- Reading slope, intercepts and perpendicular distance straight off the general equation \(Ax+By+C=0\).
- The angle between two lines, with the clean tests for parallel and perpendicular.
- The distance of a point from a line and the distance between two parallel lines.
- Intersection, concurrency, the family of lines, and the foot of perpendicular and reflection of a point.
Slope & Inclination of a Line
The inclination of a line is the angle \(\theta\), measured anticlockwise from the positive \(x\)-axis to the line, with \(0^{\circ}\le\theta<180^{\circ}\). Its slope (or gradient) is the tangent of that angle, \(m=\tan\theta\) — a single number that captures the line's steepness and direction. Through any two points the slope is the ratio of vertical change to horizontal change.
A line rising to the right has \(m>0\); falling to the right, \(m<0\). A horizontal line has \(m=0\), and a vertical line has undefined slope (\(\theta=90^{\circ}\)).
A vertical line \(x=a\) cannot be written as \(y=mx+c\) because its slope is undefined, not infinite. Keep this exception in mind whenever you divide by a difference of \(x\)-coordinates or by \(1+m_1m_2\).
Forms of a Line's Equation
A straight line is fixed by the right pair of facts — a point and a slope, or two points, or a slope and an intercept. Each pairing gives a ready-made form. They are all the same line dressed differently; choose the one that matches the data you are handed.
The point-slope form is the workhorse: it needs just one point and the slope. The two-point form is the same thing after substituting \(m=\dfrac{y_2-y_1}{x_2-x_1}\).
The line through \((2,3)\) with slope \(2\) is \(y-3=2(x-2)\), i.e. \(y=2x-1\). One point, one slope, one line.
Intercept & Normal Forms
Two more forms describe a line by where it sits rather than by a point and slope. The intercept form uses the two axis crossings; the normal form uses the perpendicular dropped from the origin.
In the intercept form, \(a\) and \(b\) are the \(x\)- and \(y\)-intercepts. In the normal form, \(p>0\) is the length of the perpendicular from the origin and \(\alpha\) the angle that perpendicular makes with the positive \(x\)-axis.
The General Equation \(Ax+By+C=0\)
Every straight line — including vertical ones — can be written as \(Ax+By+C=0\) with \(A,B\) not both zero. This single form is the one to convert into for slope, intercepts and distance, since each reads off the coefficients.
The slope is \(m=-\dfrac{A}{B}\) (provided \(B\neq0\)); if \(B=0\) the line is vertical.
\(x\)-intercept \(-\dfrac{C}{A}\), \(y\)-intercept \(-\dfrac{C}{B}\).
Divide through by \(\pm\sqrt{A^2+B^2}\) to reach the normal form and read off \(p\).
Angle Between Two Lines
Two lines with slopes \(m_1\) and \(m_2\) meet at an angle \(\theta\) found from their slopes. The acute angle is taken by the absolute value; the supplementary obtuse angle is its partner.
Parallel lines have equal slopes, \(m_1=m_2\) (then \(\tan\theta=0\)). Perpendicular lines satisfy \(m_1m_2=-1\) (the denominator vanishes, so \(\theta=90^{\circ}\)).
For lines \(A_1x+B_1y+C_1=0\) and \(A_2x+B_2y+C_2=0\): they are parallel when \(A_1B_2=A_2B_1\) and perpendicular when \(A_1A_2+B_1B_2=0\). These avoid slopes entirely, so they handle vertical lines without a special case.
Distance of a Point from a Line
The (perpendicular) distance from a point to a line is the shortest of all distances from that point to points on the line. Put the line in general form, then substitute the point's coordinates into the left side and normalise.
The first gives the distance of \((x_1,y_1)\) from \(Ax+By+C=0\). The second, for two parallel lines written with the same \(A,B\), is just the gap in their constant terms, normalised.
Intersection, Concurrency & the Family of Lines
Two non-parallel lines meet in exactly one point — solve their equations simultaneously to find it. Three or more lines are concurrent when they all pass through a single common point. And every line through a fixed intersection can be captured by a single parameter.
Three lines are concurrent precisely when this determinant vanishes — a single check that replaces finding an intersection and substituting into the third line.
Points \((x_1,y_1)\) and \((x_2,y_2)\) lie on the same side of \(Ax+By+C=0\) when \((Ax_1+By_1+C)\) and \((Ax_2+By_2+C)\) have the same sign, and on opposite sides when the signs differ.
Reflections & Standard Results
Dropping a perpendicular from a point onto a line gives its foot; reflecting the point across the line gives its image. Both come from one compact formula, and a few recurring results are worth carrying ready-made.
This locates the foot of the perpendicular from \((x_1,y_1)\) to \(Ax+By+C=0\). For the image (reflection), replace the right side with \(-\dfrac{2(Ax_1+By_1+C)}{A^2+B^2}\) — the foot is the midpoint of the point and its image.
A line parallel to \(Ax+By+C=0\) keeps \(A,B\) and only changes the constant: \(Ax+By+k=0\).
A line perpendicular to \(Ax+By+C=0\) swaps and negates: \(Bx-Ay+k=0\).
Three points are collinear when the triangle they form has zero area: \(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0\).
Putting It to Work
Problem. Find the equation of the line through \((2,3)\) with slope \(2\).
Solution. Apply the point-slope form \(y-y_1=m(x-x_1)\):
Problem. Find the equation of the line through \((1,2)\) and \((3,6)\).
Solution. The slope is \(m=\dfrac{6-2}{3-1}=2\); then point-slope through \((1,2)\):
Problem. A line makes an \(x\)-intercept of \(4\) and a \(y\)-intercept of \(3\). Find its equation.
Solution. Use the intercept form \(\dfrac{x}{a}+\dfrac{y}{b}=1\) with \(a=4,\ b=3\):
Problem. Find the acute angle between \(y=2x+1\) and \(y=-3x+4\).
Solution. With \(m_1=2,\ m_2=-3\):
Problem. Find the distance of \((3,-5)\) from the line \(3x-4y-26=0\).
Solution. Substitute into \(d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}\):
Problem. Find the foot of the perpendicular from the origin \((0,0)\) to \(x+y-2=0\), and the image of the origin in that line.
Solution. Here \(A=1,\ B=1,\ C=-2\) and \(Ax_1+By_1+C=-2\), so \(-\dfrac{-2}{1+1}=1\) for the foot and twice that for the image:
Chapter Summary
\(m=\tan\theta=\dfrac{y_2-y_1}{x_2-x_1}\); vertical lines have no slope.
Point-slope, two-point, slope-intercept, intercept \(\tfrac{x}{a}+\tfrac{y}{b}=1\), and normal \(x\cos\alpha+y\sin\alpha=p\).
\(Ax+By+C=0\) gives slope \(-A/B\) and intercepts at a glance.
\(\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|\); parallel \(m_1=m_2\), perpendicular \(m_1m_2=-1\).
\(d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}\); parallel gap \(\dfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}\).
\(L_1+\lambda L_2=0\) through an intersection; concurrency by a zero determinant.
Problems
Sketch the situation first, choose the form that matches the data, then compute. Difficulty rises down the list.
- Find the slope of the line through \((-1,2)\) and \((3,10)\).
- Write the equation of the line through \((0,4)\) with slope \(-\tfrac{1}{2}\).
- Find the equation of the line through \((2,-1)\) and \((5,8)\).
- A line cuts off intercepts \(-3\) and \(6\) on the axes. Find its equation in general form.
- Reduce \(\sqrt3\,x+y-8=0\) to normal form and state \(p\) and \(\alpha\).
- Find the acute angle between \(x-2y+3=0\) and \(3x+y-5=0\).
- Find the equation of the line through \((1,1)\) perpendicular to \(2x-3y+5=0\).
- Find the distance of \((2,3)\) from the line \(5x-12y+9=0\).
- Find the distance between the parallel lines \(3x+4y+7=0\) and \(3x+4y-3=0\).
- Find the point of intersection of \(2x+y=5\) and \(x-y=1\), and the line through it parallel to \(x+2y=0\).
- Show that \(x+y-3=0\), \(2x-y=0\) and \(x-2y+3=0\) are concurrent, and find the common point.
- Find the image of the point \((1,2)\) in the line \(3x-4y+5=0\).