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Straight LinesIntroduction

Reading time: ~15 min

Geometry, as a logical system, is a means and even the most powerful means to make children feel the strength of the human spirit that is of their own spirit. — H. FREUDENTHAL

Aman's colony is getting solar street lamps. The electrician has already put up two poles, A and B, and wants one straight overhead cable from A to B that also passes through a third pole, C. If C stands even slightly off the line, the cable bends at C, pulls on the pole, and snaps in the first monsoon storm.

On the grid below, one square is one metre. You choose where pole C goes. The shaded triangle ABC shows how far the three poles are from standing in one straight row.

Drag pole C and complete both missions.

Watch the area: both times it fell to exactly 0. Three points lie on one straight line precisely when the triangle they form has no area. A question about a picture — is the cable straight? — was answered by a formula that gave 0. Pictures answered by algebra: that is the whole idea of this chapter.

You have been doing this since Class 9. In two-dimensional coordinate geometry, every point of the plane is a pair of , and a figure can be described by an . The French philosopher and mathematician René Descartes was the first to develop this systematically, in his book La Géométrie, published in 1637: he described curves by equations and then studied them with the tools of algebra. Geometry done this way is called analytical geometry. In earlier classes you built its basic toolkit — the coordinate axes and the coordinate plane, plotting points, the distance between two points and the section formula.

Let's dust that toolkit off: four formulas from earlier classes will do a lot of work in this chapter. Fig 9.1 shows the points (6, -4) and (3, 0).

The point (6, -4) sits 6 units from the y-axis, measured along the x-axis, and 4 units from the x-axis, measured along the negative y-axis. The point (3, 0) sits 3 units from the y-axis and lies on the x-axis itself, so its distance from the x-axis is .

I. Distance formula. The distance between P(x_1, y_1) and Q(x_2, y_2) is

PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

In Fig 9.1 the two orange gaps are 3 and 4 units, so the purple segment is \sqrt{(3 - 6)^2 + (0 + 4)^2} = \sqrt{9 + 16} = units long.

II. Section formula. The point that divides the line segment joining (x_1, y_1) and (x_2, y_2) in the ratio m : n is

\left(\dfrac{m x_2 + n x_1}{m + n}, \dfrac{m y_2 + n y_1}{m + n}\right)

For A(1, -3) and B(-3, 9) in the ratio 1 : 3, this gives x = \dfrac{1 \cdot (-3) + 3 \cdot 1}{1 + 3} = 0 and y = \dfrac{1 \cdot 9 + 3 \cdot (-3)}{1 + 3} = , so the dividing point is the origin.

III. Mid-point. Put m = n in the section formula and you get the mid-point of the segment, \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right).

IV. Area of a triangle. The triangle with vertices (x_1, y_1), (x_2, y_2) and (x_3, y_3) has area

\dfrac{1}{2}\big|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\,\big|

This is the formula that ran the lamp-post widget. For the vertices (4, 4), (3, -2) and (-3, 16) it gives

\dfrac{1}{2}\big|\,4(-2 - 16) + 3(16 - 4) + (-3)(4 + 2)\,\big| = \dfrac{1}{2}\,\big|-72 + 36 - 18\big| = \dfrac{54}{2} = square units.

Note

If the area of triangle ABC is , the three points A, B and C lie on one line — they are collinear. The lamp posts showed why: three points in a straight row cannot fence in any region, so their "triangle" has flattened to nothing.

In this chapter we use these tools to study the figure in geometry: the straight line. Simple as it is, the line is everywhere — roads, rails, ramps, the edge of a ruler, the path of a light ray. Our aim is to describe any line by an equation, and the key to every one of those equations is a single number called the slope of the line.

Sina