Question

find the ratio in which the line segment joining( - 2 ,- 3) and (5 ,6) is divided x axis y axis also the co-ordinate of the point is division by in each cas
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Answer 1

Answer:

$x-\mathrm{axis\:interception\:points\:of\:}\frac{9}{7}x-\frac{3}{7}:\quad \left(\frac{1}{3},\:0\right)\\y-\mathrm{axis\:interception\:point\:of\:}\frac{9}{7}x-\frac{3}{7}:\quad \left(0,\:-\frac{3}{7}\right)$

Step-by-step explanation:

Given points of line segment terminal are (-2,-3) and (5,6)

so the equation of the line segment

$\mathrm{Find\:the\:line\:}\mathbf{y=mx+b}\mathrm{\:passing\:through\:}\left(-2,\:-3\right)\mathrm{,\:}\left(5,\:6\right)\\\mathrm{Compute\:the\:slope\:}\left(-2,\:-3\right),\:\left(5,\:6\right):\quad m=\frac{9}{7}\\\mathrm{Compute\:the\:}y\mathrm{\:intercept}:\quad b=-\frac{3}{7}\\\mathrm{Construct\:the\:line\:equation\:}\mathbf{y=mx+b}\mathrm{\:where\:}\mathbf{m}=\frac{9}{7}\mathrm{\:and\:}\mathbf{b}=-\frac{3}{7}\\\\7y=9x-3\\hence at x=0 y=-3/7\\and at y=0 x=3/9=1/3$

Hence

please check figure of line segment .

Answer 2

we know that equation of line

y = mx + c

where c is constant

$m = \frac{y2 - y1}{x2 - x1} \\ m = \frac{6 - ( - 3)}{5 - ( - 2)} \\ m = \frac{9}{7}$

y = (9/7)x + c

when x = 5 then y = 6

6 = (9/7)×5 + c

=> 42 = 45 + 7c

=> c = -3/7

equation is

y = (9/7)x - 3/7

or

7y = 9x - 3

x axis interception means y = 0

=> 0 = 9x -3

=> x = 3/9 = 1/3

y axis interception means x = 0

=> 7y = -3

=> y = -3/7

y axis intercepted at (0, -3/7)

Distance from (-2,-3)

= $\sqrt{(-2-0)^{2} + (-3 -(-3/7)^2} = \sqrt{(-2)^2 + (\frac{-18}{7})^2  } \\= \sqrt{4 + \frac{324}{49} } = \sqrt{\frac{520}{49}}$

Distance from (5, 6)

= $\sqrt{(5-0)^{2} + (6 -(-3/7)^2} = \sqrt{(5)^2 + (\frac{45}{7})^2  } \\= \sqrt{25 + \frac{2025}{49} } = \sqrt{\frac{3250}{49}}$

Ratio = $\sqrt{\frac{3250}{49}} / \sqrt{\frac{520}{49}} = \sqrt{6.25} = 2.5$

x axis interception at (1/3 , 0)

Distance from (-2,-3) = $\sqrt{(-2- \frac{1}{3})^2 + (-3-0)^2} = \sqrt{\frac{49}{9} + 9} = \sqrt{\frac{130}{9}}$

Distance from (5,6) =  $\sqrt{(5- \frac{1}{3})^2 + (6-0)^2} = \sqrt{\frac{196}{9} + 36} = \sqrt{\frac{520}{9}}$

Ratio = $\sqrt{\frac{520}{130}} = \sqrt{4} = 2$