Question

Find the ratio in which the line segment joining A( 1 , 5 ) and B( - 4 , 5 ) is divided by the x- axis. Also find the coordinates of point of division.

Answer 1

☯ Let the ratio in which line segment joining A and B is divided by the x- axis be k:1.

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$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(3.5,5){\line(1,0){8}}\put(3,4.5){\bf A}\put(11.5,4.5){\bf B}\put(6.5,4.4){\bf P}\put(6.7,4.8){\line(0,1){0.4}}\put(2.5,3.8){\sf (1, - 5)}\put(11,3.8){\sf ( - 4,5)}\put(6.2,3.8){\sf (x, 0)}\put(5,5.5){\bf k}\put(8,5.5){\bf 1}\end{picture}$

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Here,

• Point P lies on x axis, hence its y cordinate is 0. So, P(x,0).

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$\underline{\bigstar\:\boldsymbol{Using\:section\:formula\::}}$

$\star\;{\boxed{\sf{\pink{(x,y) = \bigg( \dfrac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\;,\; \dfrac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \bigg)}}}}$

$\sf where \begin{cases} & \sf{m_1 = k\;,\;m_2 = 1 } \\ & \sf{x_1 = 1\;,\; y_1 = -5} \\ & \sf{x_2 = -4\;,\; y_2 = 5} \\ & \sf{x = x\;,\;y = 0} \end{cases}$

$:\implies\sf (x,0) = \bigg( \dfrac{k \times (-4) + 1 \times 1}{k + 1}\;,\; \dfrac{k \times 5 + 1 \times (-5)}{k + 1} \bigg)$

$:\implies\sf (x,0) = \bigg( \dfrac{-4k + 1}{k + 1}\;,\; \dfrac{5k - 5}{k + 1} \bigg)$

Therefore,

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$:\implies\sf 0 = \dfrac{5k - 5}{k + 1}\\ \\ :\implies\sf 0 \times (k + 1) = 5k - 5\\ \\ :\implies\sf 5k - 5 = 0\\ \\ :\implies\sf 5k = 5\\ \\ :\implies\sf k = \cancel{ \dfrac{5}{5}}\\ \\ :\implies{\underline{\boxed{\sf{\purple{k = 1}}}}}\;\bigstar$

Now, We need to find x also,

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$:\implies\sf x = \dfrac{-4(1) + 1}{1 + 1}\\ \\ :\implies\sf x = \dfrac{-4 + 1}{2}\\ \\ :\implies{\underline{\boxed{\sf{\purple{x = \dfrac{-3}{2}}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{The\; cordinate\;of\;point\;P(x,0)\;is\; \bf{ \dfrac{-3}{2},0}.}}}$

Answer 2

Let the ratio in which line segment joining A and B is divided by the x- axis be k:1.

Here,

Point P lies on x axis, hence its y cordinate is 0. So, P(x,0).

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★Usingsectionformula:

⋆ (x,y)=( m 1 +m 2m 1 x 2 +m 2 x x1 , m 1 +m 2m 1 y 2 +m 2 y y1 )m 1

=k,m 2

=1x 1

=1,y 1

=−5x 2

=−4,y

2 =5

=x

y=0

:⟹(x,0)=( k+1k×(−4)+1×1 , k+1k×5+1×(−5) )

:⟹(x,0)=( k+1−4k+1 , k+15k−5 )

Therefore,

:⟹0= k+15k−5

:⟹0×(k+1)=5k−5

:⟹5k−5=0

:⟹5k=5

:⟹k= 5/5

:⟹ k=1

★Now, We need to find x also,

:⟹x= 1+1−4(1)+1

:⟹x= 2−4+1

:⟹ x= 2−3

★ThecordinateofpointP(x,0)is 2−3 ,0.