Slope and angle between lines

Slope from two points. A line is tilted the same way everywhere, so any two of its points give the same ratio of rise to run. For a line through $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1 \ne x_2$,

where: $m$ is the slope (a pure number, no units), and $(x_1,y_1)$, $(x_2,y_2)$ are any two distinct points on the line, subtracted in matching order.

Slope as inclination. Every non-vertical line makes a definite angle $\theta$ with the positive $x$-axis, measured anticlockwise, $0° \le \theta < 180°$. The slope is simply $m=\tan\theta$. A horizontal line has $\theta=0°$, so $m=0$; a vertical line has $\theta=90°$, where tangent is undefined, so vertical lines have no defined slope — they are the one case slope cannot describe.

Reading the sign. A positive slope means the line climbs as you move right (acute inclination); a negative slope means it falls (obtuse inclination); zero slope means flat. This sign picture is exactly why the angle-between-lines formula below needs an absolute value — two lines of opposite-sign slope meet at both an acute and an obtuse angle, and we always want the acute one first.

Angle between two lines. For two lines with slopes $m_1$ and $m_2$ (neither vertical), the acute angle $\theta$ between them satisfies

where: $m_1, m_2$ are the two slopes and the modulus guarantees $\theta$ lands in $[0°,90°)$, the acute angle; the obtuse angle between the same pair of lines is $180°-\theta$.

Parallel and perpendicular conditions. Two non-vertical lines are parallel exactly when $m_1=m_2$ (two vertical lines are also parallel to each other, despite having no slope). Two non-vertical, non-horizontal lines are perpendicular exactly when

equivalently $m_2=-1/m_1$, each slope the negative reciprocal of the other. A horizontal and a vertical line are perpendicular by direct inspection, not by this formula, since the vertical one has no slope to plug in.

Find the angle between the lines with slopes $m_1=3$ and $m_2=-2$, and then find the equation of the line through $(2,1)$ that is perpendicular to the line of slope $m_1=3$.

Step 1. Substitute into the angle formula: $\tan\theta=\left|\dfrac{3-(-2)}{1+(3)(-2)}\right|=\left|\dfrac{5}{1-6}\right|=\left|\dfrac{5}{-5}\right|=1$.

Step 2. Solve for the angle: $\theta=\tan^{-1}(1)=45°$. This is the acute angle between the two lines; the obtuse angle between the same pair is $180°-45°=135°$.

Step 3. For the perpendicular line, its required slope $m_3$ must satisfy $m_1 m_3=-1$ with $m_1=3$, so $m_3=-\dfrac{1}{3}$.

Step 4. Apply point-slope form through $(2,1)$ with slope $-\dfrac13$: $y-1=-\dfrac13(x-2)$, which simplifies to $3y-3=-x+2$, or $x+3y-5=0$.

The acute angle between the two given lines is $\boxed{45°}$, and the perpendicular line through $(2,1)$ is $x+3y-5=0$.

Competitive stretch

A line $\ell$ passes through $A(1,2)$ and $B(4,8)$, and a second line $m$ passes through $C(-2,-1)$ and $D(0,3)$. Show $\ell$ is parallel to $m$, then find a point $E$ with $x$-coordinate $7$ so that segment $AE$ is perpendicular to $\ell$.

Step 1. Slope of $\ell=AB$: $m_1=\dfrac{8-2}{4-1}=\dfrac{6}{3}=2$. Slope of $m=CD$: $m_2=\dfrac{3-(-1)}{0-(-2)}=\dfrac{4}{2}=2$. Since $m_1=m_2=2$, the two lines are parallel, and since they clearly pass through different points, they are distinct parallel lines, not the same line.

Step 2. The line $AE$ must be perpendicular to $\ell$, whose slope is $2$, so the required slope $m_3$ satisfies $2m_3=-1$, giving $m_3=-\dfrac12$.

Step 3. Apply point-slope form through $A(1,2)$ with slope $-\dfrac12$: $y-2=-\dfrac12(x-1)$. At $x=7$: $y-2=-\dfrac12(6)=-3$, so $y=-1$.

$\ell$ and $m$ are parallel (both slope $2$), and $E=(7,-1)$ makes $AE$ perpendicular to $\ell$.

• Dividing coordinates in the wrong order. Slope must be $\dfrac{y_2-y_1}{x_2-x_1}$ with both differences taken in the same point order — swapping one but not the other flips the sign.
• Dropping the absolute value in the angle formula. Without it, $\tan\theta$ can come out negative, which does not correspond to any angle in the expected acute range — always take the modulus first.
• Applying $m_1m_2=-1$ to a vertical or horizontal pair. A vertical line has no slope, so it cannot be substituted into this formula; a horizontal and a vertical line are perpendicular by inspection instead.
• Mixing up parallel with coincident. Equal slopes only guarantee the lines are parallel, not identical — check the intercepts (or a shared point) if the question asks whether two lines are actually the same line.

Ask the tutor: If you get a negative value under the square-root-free tangent formula and are unsure whether to keep or flip the sign, ask the AI tutor to walk through it with your specific pair of slopes.