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Official Sr. Teacher Gr II NON-TSP MATHEMATICS (Held on :29 Oct 2018)

Option 2 : \(\frac 5 2\)

Army Havildar SAC 2021 Full Mock Test

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100 Questions
200 Marks
120 Mins

**Concept:**

If three points (x1, y1), (x2, y2) and (x3, y3) are** collinear** then the area of the triangle determined by the three points is zero.

\(\left| {\begin{array}{*{20}{c}} {{{\rm{x}}_1}}&{{{\rm{y}}_1}}&1\\ {{{\rm{x}}_2}}&{{{\rm{y}}_2}}&1\\ {{{\rm{x}}_3}}&{{{\rm{y}}_3}}&1 \end{array}} \right|{\rm{\;}} = {\rm{\;}}0\)

or

If three or more points are **collinear** then the slope of any two pairs of points is same.

For example, let three points A, B and C are collinear then

slope of AB = slope of BC = slope of AC

Slope of the line if two points \(({{\rm{x}}_1},{{\rm{y}}_1}){\rm{\;and\;}}\left( {{{\rm{x}}_2},{\rm{\;}}{{\rm{y}}_2}} \right)\) are given by:

\(\Rightarrow \left( {\bf{m}} \right) = \;\frac{{{{\bf{y}}_2} - {{\bf{y}}_1}}}{{{{\bf{x}}_2} - {{\bf{x}}_1}}}{\rm{\;}}\)

**Calculation:**

Given: the points (-2, -5), (2, -2) and (8, a) are collinear

\(\begin{vmatrix} -2 &-5 &1 \\ 2 &-2 &1 \\ 8& \rm a &1 \end{vmatrix} = 0 \\\Rightarrow -2{(-2-\rm a)}-(-5)(2-8)+1(2a+16)=0\\\Rightarrow4+2\rm a-30+2\rm a + 16 = 0\\\Rightarrow 4\rm a-10=0\\\Rightarrow 4\rm a = 10\\\therefore \rm a = \frac{5}{2}\)