Show that the points (1, 3), (2, 1) and (dfrac12, 4) are collinear, by using (i) the concept of slope, (ii) th

Exercise 6.2 · Q10

(1, 3)

(i) the concept of slope,

(ii) the equation of a straight line, and

(iii) any other method.

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Concept understanding — Straight Lines — Forms of the Equation

The general (linear) equation of a straight line is $ax+by+c=0$ , where $a,b$ are not both zero; the set of solutions of any such equation is a straight line in the plane. Because dividing through by $b$ (or $a$ ) removes one constant, every line's equation genuinely contains only two independent arbitrary constants — so exactly two independent pieces of information (two points, or a point and a slope, or two intercepts, etc.) are enough to pin a line down uniquely.

Slope. The angle of inclination $\theta$ of a line is the angle it makes with the $x$ -axis, measured counter-clockwise; the slope $m=\tan\theta$ (undefined when $\theta=\pi/2$ , i.e. for a vertical line). Equivalently, through two points $(x_1,y_1),(x_2,y_2)$ with $x_1\ne x_2$ , $m=\dfrac{y_2-y_1}{x_2-x_1}$ ; from the general form, $m=-a/b$ ( $b\ne0$ ). Three points are collinear exactly when the slope of any one pair equals the slope of another pair sharing a point.

Intercepts. The $x$ -intercept is where a line meets the $x$ -axis ( $y=0$ ); the $y$ -intercept is where it meets the $y$ -axis ( $x=0$ ). ( $x=0$ is itself the equation of the $y$ -axis; $y=0$ is the equation of the $x$ -axis.)

The six forms (two conditions each, all interconvertible by algebra):

Data given	Equation
Slope $m$ , $y$ -intercept $b$	$y=mx+b$
Slope $m$ , point $(x_1,y_1)$	$y-y_1=m(x-x_1)$
Two points $(x_1,y_1),(x_2,y_2)$	$\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}$ , equivalently $\begin{vmatrix}x-x_1 & y-y_1\\ x_2-x_1 & y_2-y_1\end{vmatrix}=0$
$x$ -intercept $a$ , $y$ -intercept $b$ (both $\ne0$ )	$\dfrac{x}{a}+\dfrac{y}{b}=1$
Normal length $p$ , angle $\alpha$ of the normal with the $x$ -axis	$x\cos\alpha+y\sin\alpha=p$
Parametric, through $(x_1,y_1)$ at inclination $\theta$ , parameter $r$ = signed distance from $(x_1,y_1)$	$\dfrac{x-x_1}{\cos\theta}=\dfrac{y-y_1}{\sin\theta}=r$

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Verify collinearity of $(1,3),(2,1),\left(\tfrac12,4\right)$ three independent ways: equal slopes, the third point satisfying the line through the first two, and zero triangle area.

Let $A(1,3)$ , $B(2,1)$ , $C\left(\dfrac12,4\right)$ .

Step 1. Method (i) — compare slopes. Slope of $AB$ :

m_{AB}=\frac{1-3}{2-1}=\frac{-2}{1}=-2

Slope of $BC$ :

m_{BC}=\frac{4-1}{\frac12-2}=\frac{3}{-\frac32}=-2

Since $B$ is common to both pairs and $m_{AB}=m_{BC}=-2$ , the three points lie on the same straight line, so $A,B,C$ are collinear.

Step 2. Method (ii) — equation of the line through two points, then test the third. Line through $A(1,3)$ and $B(2,1)$ , slope $-2$ :

y-3=-2(x-1) \;\Rightarrow\; y-3=-2x+2 \;\Rightarrow\; 2x+y=5

Test $C\left(\dfrac12,4\right)$ : $2\left(\dfrac12\right)+4=1+4=5$ ✓ — $C$ lies on the same line, confirming collinearity.

Step 3. Method (iii) — area of the triangle formed by the three points. Using

$\text{Area}=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ …

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Q.Show that the points (1,3)(1, 3)(1,3), (2,1)(2, 1)(2,1) and (12,4)\left(\dfrac12, 4\right)(21​,4) are collinear, by using

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Q.Show that the points $(1, 3)$ , $(2, 1)$ and $\left(\dfrac12, 4\right)$ are collinear, by using

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