Find the ratio in which the …

we know that equation of line

y = mx + c

where c is constant

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

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)

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

Distance from (5, 6)

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

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

x axis interception at (1/3 , 0)

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

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

{tex}Ratio = \sqrt{\frac{520}{130}} = \sqrt{4} = 2{/tex}