Reading a line at sight
Tonight's idea
A line crosses the \(y\)-axis where \(x = 0\) and the \(x\)-axis where \(y = 0\). Those crossing points are its intercepts. In \(y = mx + b\), put \(x = 0\): \(y = b\), so the \(y\)-intercept is read straight off. For the \(x\)-intercept, put \(y = 0\) and solve \(mx + b = 0\).
Lines also arrive as \(2x + 3y = 6\). Do not rearrange. Cover the \(y\) term: \(2x = 6\), \(x = 3\). Cover the \(x\) term: \(3y = 6\), \(y = 2\). Two points: the line is drawn.
Two lines are parallel when they have the same slope and different intercepts. Same slope means they climb at the same rate, so the gap never changes. In the figure, both lines rearrange to slope \(-\tfrac23\): \(y = -\tfrac23 x + 2\) and \(y = -\tfrac23 x + 4\). Same \(m\), different \(b\).
Two special cases. A horizontal line has slope 0: \(y = 0 \cdot x + b\), written \(y = b\). A vertical line has run 0, and dividing by 0 is not allowed, so its slope is undefined and it cannot be written as \(y = mx + b\). Its equation is \(x = c\): every point on it has the same \(x\).
From two points: find \(m\) as rise over run, then substitute one point to find \(b\). No formula needed.
Watch one: the line through \((1, 5)\) and \((4, -1)\)
Run \(3\), rise \(-6\), so \(m = -2\). Then \(y = -2x + b\) with \((1, 5)\) on the line: \(5 = -2 + b\), so \(b = 7\). The line is \(y = -2x + 7\).
Intercepts: \(y\)-intercept 7. \(x\)-intercept: \(0 = -2x + 7\), so \(x = 3.5\).
Is it parallel to \(4x + 2y = 9\)? Rearrange: \(2y = -4x + 9\), so \(y = -2x + 4.5\). Same slope, different intercept: yes. Parallel to \(2x - y = 5\)? That is \(y = 2x - 5\), slope 2, not \(-2\). No.
Slide \(m\) and \(b\) and switch on the parallel ghost line: mathdetours.com/demos/lines.html.
Your turn
- Find both intercepts of each line: (a) \(3x + 4y = 12\); (b) \(y = -2x + 5\); (c) \(2x - 5y = 20\).
- Write the equation of: (a) the horizontal line through \((3, -2)\); (b) the vertical line through \((3, -2)\), and say why it has no slope; (c) the line through \((3, -2)\) parallel to \(y = 4x - 1\).
- A line passes through \((-2, 1)\) and \((4, 10)\). (a) Find its equation. (b) Find its intercepts. (c) Which of these are parallel to it: \(y = 1.5x - 7\); \(3x - 2y = 8\); \(2x - 3y = 8\); \(y = 4 + 1.5x\)? One of them is a trap.
- A line crosses the \(x\)-axis at 4 and the \(y\)-axis at \(-6\). (a) Write its equation. (b) A parallel line crosses the \(y\)-axis at 2. Find where it crosses the \(x\)-axis, first from its equation, then with no equation: what do the two triangles cut off by the axes have in common?
Stretch: the intercept form
A line has \(x\)-intercept \(a\) and \(y\)-intercept \(b\), both non-zero. Show that its equation can be written \(\dfrac{x}{a} + \dfrac{y}{b} = 1\). Then, still with non-zero intercepts, find the line through \((2, 3)\) whose two intercepts are equal, and the line through \((2, 3)\) whose \(y\)-intercept is double its \(x\)-intercept.
Answers
- (a) \((4, 0)\) and \((0, 3)\). (b) \((2.5, 0)\) and \((0, 5)\). (c) \((10, 0)\) and \((0, -4)\).
- (a) \(y = -2\). (b) \(x = 3\). Any two points on it have run 0, and rise ÷ 0 is undefined. (c) Slope 4: \(y = 4x + b\) with \(-2 = 12 + b\), so \(y = 4x - 14\).
- (a) \(m = 9/6 = \tfrac32\); \(1 = \tfrac32(-2) + b\) gives \(b = 4\), so \(y = \tfrac32 x + 4\). (b) \((0, 4)\) and \((-\tfrac83, 0)\). (c) \(y = 1.5x - 7\): parallel. \(3x - 2y = 8\) is \(y = 1.5x - 4\): parallel. \(2x - 3y = 8\) is \(y = \tfrac23 x - \tfrac83\): not parallel. \(y = 4 + 1.5x\) is the same line, not a parallel one. That is the trap.
- (a) From \((0, -6)\) to \((4, 0)\): rise 6, run 4, \(m = \tfrac32\), so \(y = \tfrac32 x - 6\). (b) \(y = \tfrac32 x + 2\); at \(y = 0\), \(x = -\tfrac43\). Without the equation: each line cuts a right triangle off the axes, and the hypotenuses have the same slope, so the triangles are similar. Legs \(4\) and \(6\) are in ratio \(2 : 3\), so a \(y\)-leg of 2 goes with an \(x\)-leg of \(\tfrac43\). The line rises to the right and is already above the origin at \(x = 0\), so it crossed the \(x\)-axis to the left: at \(-\tfrac43\).
Stretch. From \((a, 0)\) to \((0, b)\) the slope is \(\dfrac{b - 0}{0 - a} = -\dfrac{b}{a}\), so \(y = -\dfrac{b}{a}x + b\). Divide every term by \(b\): \(\dfrac{y}{b} = -\dfrac{x}{a} + 1\), which is \(\dfrac{x}{a} + \dfrac{y}{b} = 1\). Equal intercepts \(a = b\): \(\dfrac{2}{a} + \dfrac{3}{a} = 1\), so \(a = 5\) and the line is \(x + y = 5\). With \(b = 2a\): \(\dfrac{2}{a} + \dfrac{3}{2a} = \dfrac{7}{2a} = 1\), so \(a = \tfrac72\), \(b = 7\), and the line is \(2x + y = 7\). Check: \(4 + 3 = 7\).