Question

Determine whether the given set of points in each case are collinear or not.
(i) (7,–2),(5,1),(3,4) (ii) (–2, –8), (2,–3) (6,2)
(iii) (a,–2), (a,3), (a,0)

Answer 1

Answer:

(i) collinear

(ii) collinear

(ii) non collinear

Step-by-step explanation:

Since, three points $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ are called col-linear, if,

$x_1(y_2-y_3)+x_2(y_3-y_1) + x_3(y_1-y_2)=0$

(i) $x_1 = 7, y_1 = -2, x_2 = 5, y_2 = 1, x_3 = 3, y_3 = 4$

Since,

$7(1-4)+5(4+2) + 3(-2-1)=7(-3) + 5(6) + 3(-3) = -21 + 30 - 9 = 9 - 9=0$

Thus,

(7,–2),(5,1) and (3,4) are col-linear.

(ii) $x_1 = -2, y_1 = -8, x_2 = 2, y_2 = -3, x_3 = 6, y_3 = 2$

Since,

$-2(-3-2)+2(2+8) + 6(-8+3)=-2(-5) + 2(10) + 6(-5) = 10 + 20 - 30 = 30 - 30=0$

Thus,

(–2, –8), (2,–3) and (6,2) are col-linear.

(i) $x_1 = a, y_1 = -2, x_2 = a, y_2 = 3, x_3 = a, y_3 = 0$

Since,

$a(3-0)+a(0+3) + a(-2-3)=3a + 3a - 2a - 3a = a\neq 0$

Thus,

(7,–2),(5,1) and (3,4) are not col-linear.

Answer 2

Answer:

$i) Given \: points \: (7,-2),\:(5,1) \:and \: (3,4)$

$Let \: A(7,-2) = (x_{1},y_{1}), \\B(5,1) = (x_{2},y_{2}),\\C(3,4) = (x_{3},y_{3})$

$\boxed { \pink {Area \triangle =\frac{1}{2} | x_{1}(y_{2}-y_{3})+x_{2}(y_{3} - y_{1})+x_{3}(y_{1} - y_{2})|\:}}$

$= \frac{1}{2} | 7[1-4]+5[4-(-2)]+3[-2-1]|$

$= \frac{1}{2} | 7\times (-3) + 5\times 6+3\times (-3)|$

$= \frac{1}{2} | -21 + 30 -9|$

$= \frac{1}{2} | -30 + 30 |$

$= \frac{1}{2} \times 0$

$= 0$

Therefore.,

$\green { Area \: of \: triangle = 0 }$

$\green { A,B \: and \: C \: are \: collinear}$

$ii) Given \: points \: (-2,-3),\:(2,-3) \:and \: (6,2)$

$Let \: A(-2,-3) = (x_{1},y_{1}), \\B(2,-3) = (x_{2},y_{2}),\\C(6,2) = (x_{3},y_{3})$

$= \frac{1}{2} | (-2)[-3-2]+2[2-(-8)]+6[-8-(-3)]|$

$= \frac{1}{2} | (-2)\times (-5) + 2\times 10+6\times (-5)|$

$= \frac{1}{2} | 10 + 20 -30|$

$= \frac{1}{2} | -30 + 30 |$

$= \frac{1}{2} \times 0$

$= 0$

Therefore.,

$\green { Area \: of \: triangle = 0 }$

$\green { A,B \: and \: C \: are \: collinear}$

$iii) Given \: points \: (a,-2),\:(a,3) \:and \: (a,0)$

$Let \: A(a,-2) = (x_{1},y_{1}), \\B(a,3) = (x_{2},y_{2}),\\C(a,0) = (x_{3},y_{3})$

$= \frac{1}{2} | a[3-0]+a[0-(-2)]+a[-2-3]|$

$= \frac{1}{2} | a\times 3+ a\times 2+a\times (-5)|$

$= \frac{1}{2} | 3a + 3a -5a|$

$= \frac{1}{2} | 6a -5a |$

$= \frac{1}{2} \times a$

$≠0$

Therefore.,

$\green { Area \: of \: triangle ≠ 0 }$

$\green { A,B \: and \: C \: are \: non- collinear}$

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