Question

if the point (1, x) , (5, 2) and (9, 5) are collinear then the value x is​

Answer 1

Answer:-

Given:

(1 , x) , (5 , 2) & (9 , 5) are collinear points.

We know that,

If three points are collinear then the area of the triangle formed by them will be zero.

$\sf Area\: of \:a \: \triangle = \dfrac{1}{2}\begin{vmatrix}\sf x_1-x_2&\sf x_1 - x_3\\\\\sf y_1-y_2&\sf y_1-y_3 \end{vmatrix} \quad$

So, area of the ∆ formed by the given points = 0.

Let,

• x₁ = 1

• x₂ = 5

• x₃ = 9

• y₁ = x

• y₂ = 2

• y₃ = 5

Hence,

$\implies\sf Area \: of \: the \: \triangle = \dfrac{1}{2}\begin{vmatrix}\sf 1-5&\sf 1 - 9\\\\\sf \: x-2&\sf x-5 \end{vmatrix} \\ \\ \implies\sf 0 = \frac{1}{2} \begin{vmatrix} \sf \: - 4& \sf \: - 8 \\ \\ \sf \: x - 2& \sf \: x - 5 \end{vmatrix} \\ \\ \\ \\ \implies \sf \: 0 = |(x - 5)( - 4) - ( - 8)(x - 2)| \\ \\ \implies \sf \: 0 = | - 4x + 20 - ( - 8x + 16)| \\ \\ \implies \sf \: 0 = | - 4x + 20 + 8x - 16| \\ \\ \implies \sf \: 0 = |4x + 4| \\ \\ \implies \sf \: 4x = - 4 \\ \\ \implies \sf \: x = \frac{ - 4}{4} \\ \\ \implies \boxed{ \sf x = - 1}$

Therefore, the value of x is - 1.

Answer 2

$\implies \boxed{ \sf{there \: are \: two \: methods \: in \: which \: this}}</p><p>$

$\implies \boxed{ \sf{question \: can \: be \: solved}}$

${\huge{\boxed{\overline{\mid{Method \:1}}}}}}$

$\implies \boxed{ \sf{For \: the \: collinear \: all \: three \: point}} \\ \boxed{ \sf{ must \: lie \: on \: the \: same \: line.}}</p><p>$

$\implies \boxed{ \sf{let \: the \: given \: 3 points \: let}} = \\ \boxed{ \sf{ A(1,X),B(5,2),C(9,5)}}$

$\implies\boxed{ \sf{Consider \: B \: Let (x1=5,y1=2)}}$

$\implies\boxed{ \sf{Similarly \: for \: A \: C }}\\ \boxed{ \sf{(x3=1 \: y3=X) (x2=9 \: y2=5)}} \\ \boxed { \sf{ respectively.</p><p>}}$

$\implies\boxed{ \sf{Lets \: recall \: the \: the slope \: point \: form \: }} \\ \boxed{ \sf{ of straight line </p><p>}}$

$\implies\boxed{ \sf{(y-y1)=m(x-x1)}}$

$\implies\boxed{ \sf{Where \: m \: is \: slope \: which \: is \: equal \:}} \\ \boxed{ \sf{to\frac{(y2-y1)}{(x2 - x1)} }}$

$\implies \boxed{ \sf{In \: this \: question \: m= \frac{5 - 2}{9 - 5 } = \frac{3}{4} }}$

$\implies\boxed{ \sf{Now \: using \: slope \: point \: form:}}$

$\implies\boxed{ \sf{y-2=(x-5)3/4}} \\ \boxed{ \sf{ = 4y-8=3x-15}}</p><p>$

$\implies\boxed{ \sf{3x-4y-7=0}} \\ \boxed{ \sf {(this \: the \: equation \: of \: line)}}$

$\implies\boxed{ \sf{Now \: if \: we \: put \: the \: valu \: e of \: x \: }} \\ \boxed{ \sf{ co -oridinate \: then \: we \: must}}$

$\implies\boxed{ \sf{get \: the \: value \: of \: y \: co \: oridinate}}$

$\implies\boxed{ \sf{Put \: point \: A(1,X)}}$

$\implies\boxed{ \sf{Put \: point \: A(1,X)}}$

$\implies\boxed{ \sf{3 - 4X-7=0}}$

$\implies\boxed{ \sf{-4X=4}}$

$\implies\boxed{ \sf{X= -1}}$

${ \huge{ \boxed{ \overline{ \mid{Method \: 2}}}}}$

$\implies \boxed{ \sf{For \: collinear \: all \: three \: point \: must \: }} \\ \boxed{ \sf{lie \: on \: same \: line \: and \: the \: slope \: will \: be \: }} \\ \boxed{ \sf{ equal \: for \: each \: and \: every \: point.}}$

$\implies\boxed{ \sf{ \frac{(y2 - y1)}{(x2 - x1)} = \frac{(y3 - y2)}{(x3 - x1)} }}</p><p>$

$\implies \boxed{ \sf{ \frac{5 - 2}{9 - 5} = \frac{x - 5}{1 - 9} }}$

$\implies \boxed{ \sf{-6=X-5}}</p><p> \\ \implies\boxed{ \sf{X=-1}}$