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ICSE Class X Math 2014 Paper

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Question : 28 of 52
Marks: +1, -0
Find the value of ' aa ' for which the following points A(a,3),B(2,1)A(a, 3), B(2,1) and C(5,a)C(5, a) are collinear. Hence find the equation of the line.
Solution:  
Given : A(a,3),B(2,1)A(a, 3), B(2,1) and C(5,a)C(5, a) are collinear.
  ∴    Slope of   AB  =  Slope of   BC\;\therefore\;\; \text{Slope of }\; AB\;=\;\text{Slope of }\; BC
⇒    1−32−a  =  a−15−2\Rightarrow\;\; \frac{1-3}{2-a}\;=\;\frac{a-1}{5-2}
⇒    −22−a  =  a−13\Rightarrow\;\; \frac{-2}{2-a}\;=\;\frac{a-1}{3}
⇒−6  =(2−a)(a−1)\Rightarrow -6\;=(2-a)(a-1)
⇒−6  =2a−2−a2+a\Rightarrow -6\;=2a-2-a^2+a
⇒  a2−3a−4  =0\Rightarrow\;a^2-3a-4\;=0
⇒a2−4a+a−4  =0\Rightarrow a^2-4a+a-4\;=0
⇒  (a−4)(a+1)  =0\Rightarrow\;(a-4)(a+1)\;=0
  a  =4,−1\;a\;=4,-1
Now equation of straight line are :
when a=4a=4 ,
  y−3=  1−32−4(x−4)\;y-3=\;\frac{1-3}{2-4}(x-4)
  y−3=  −2−2(x−4)\;y-3=\;\frac{-2}{-2}(x-4)
y−3  =x−4y-3\;=x-4
y  =x−1y\;=x-1
when a=−1a=-1,
y−3  =  1−32+1(x+1)y-3\;=\;\frac{1-3}{2+1}(x+1)
y−3  =−  23(x+1)y-3\;=-\;\frac{2}{3}(x+1)
3y−9  =−2x−23y-9\;=-2x-2
3y+2x  =73y+2x\;=7
Hence the equations of lines are
y=x−1  and  3y+2x=7y=x-1 \;\text{and}\; 3y+2x=7
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