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 the point of division.

Answer 1

Step-by-step explanation:

Given :-

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 the point of division.

To Find :-

The Ratio

Note :-

Point P is on x-axis hence, it's y coordinate is 0. So, it is of the form P (x, 0)

Solution :-

Let Ratio be k : 1

Hence,

m1 = k , m2 = 1

x1 = 1 , x2 = -4

y1 = -5 , y2 = 5

Also,

x = x and y = 0

Using section formula,

$y = \frac{m1 \: y2 \: + \: m2 \: y1}{m1 \: + \: m2} \\ \\ 0 = \frac{k \: \times \: 5 \: + \: 1 \: x \: - \: 5}{k \: + \: 1} \\ \\ 0 = \frac{5k \: - \: 5}{k \: + \: 1} \\ \\ 0(k \: + \: 1) = 5k \: - \: 5 \\ \\ 0 = 5k \: - \: 5 \\ \\ 5k \: - \: 5 = 0 \\ \\ 5k = 5 \\ \\ k = \frac{5}{5} \\ \\ k = 1$

Hence, k = 1

Now, we need to find x.

$x = \frac{m1 \: x2 \: + \: m2 \: x1}{m1 \: + \: m2} \\ \\ = \frac{k \: x \: - \:4 \: + \: 1 \: \times 1 }{k \: + \: 1} \\ \\ = \frac{k \: x \: - \:4 \: + \: 1 \: \times 1}{1 \: + \: 1} \\ \\ = \frac{ - 4 \: + \: 1}{2} \\ \\ = \frac{ - 3}{2}$

Hence, the coordinate of point P is (x, 0)

$P = (\frac{ - 3}{2}) \: and \: 0$

Answer 2

Answer:

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

⠀⠀⠀⠀

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

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

Here,

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

⠀⠀⠀⠀

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

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

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

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

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

Therefore,

⠀⠀⠀⠀

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

Now, We need to find x also,

⠀⠀⠀⠀

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

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