Q.Using slopes, show that the points $(4, 4)$, $(3, 5)$ and $(-1, -1)$ are the vertices of a right-angled triangle, and state at which vertex the right angle occurs.

Angle Between Two Lines – From Intuition to Precision

When you think of two lines crossing each other, the first thing you notice is how "wide" or "narrow" the opening between them is. That opening is the angle between the lines. If you hold two pens and let them cross, the smaller turn you make to bring one pen onto the other is the angle between them.

But here's the key: two intersecting lines actually make four angles — two acute (sharp) and two obtuse (wide), or all four right angles if they are perpendicular. By convention, when we say "the angle between two lines," we always mean the smaller (acute) angle, which lies between $0^\circ$ and $90^\circ$. If the lines are parallel, the angle is $0^\circ$; if they are perpendicular, it is $90^\circ$.

The Geometry of Slopes

Every non-vertical line in the coordinate plane has a slope $m$, which tells you how steep it is. The slope is the tangent of the angle the line makes with the positive $x$-axis. So if a line makes an angle $\theta$ with the $x$-axis, then $m = \tan \theta$.

Now imagine two lines with slopes $m_1$ and $m_2$. They make angles $\theta_1$ and $\theta_2$ with the $x$-axis. The angle between the lines themselves is simply the difference between these two angles: $|\theta_1 - \theta_2|$.

Here $\phi$ is the acute angle between the two lines. The absolute value ensures we get the smaller angle. The denominator $1 + m_1 m_2$ comes from the tangent subtraction formula: $\tan(\theta_1 - \theta_2) = \frac{\tan \theta_1 - \tan \theta_2}{1 + \tan \theta_1 \tan \theta_2}$.

Why the Formula Works

Suppose line $L_1$ has slope $m_1 = \tan \theta_1$ and line $L_2$ has slope $m_2 = \tan \theta_2$. The angle between them is $\phi = |\theta_1 - \theta_2|$. Using the tangent subtraction identity:

The absolute value guarantees we take the acute angle. If $1 + m_1 m_2 = 0$, the denominator is zero, meaning $\tan \phi$ is undefined — that happens when $\phi = 90^\circ$, i.e., the lines are perpendicular.

Special Cases

• Parallel lines: $m_1 = m_2$. Then numerator is zero, so $\tan \phi = 0$, giving $\phi = 0^\circ$.
• Perpendicular lines: $m_1 m_2 = -1$. Then denominator is zero, so $\phi = 90^\circ$. …