Question

find the ratio in which y axis divid the line segment joining the points A(5,-6)and(-1,-4)?and also find the point of intersection. ​

Answer 1

Step-by-step explanation:

Let P(0, b) divides the line segment joining the points A(5,-6) and B(-1,-4)in the ratio m : n.

Therefore, by section formula:

$x = \frac{mx_2+nx_1}{m+n}\\</p><p>\implies 0=\frac{m\times (-1)+n\times (5)}{m+n}\\</p><p>\implies 0 \times (m+n) = - m + 5n\\</p><p>\implies 0 = - m + 5n\\</p><p>\implies m = 5n\\</p><p>\implies \frac {m}{n} = \frac {5}{1}\\</p><p>\implies \huge\fbox {m : n= 5 : 1}\\\\next\\ </p><p>y = \frac{my_2+ny_1}{m+n}\\</p><p>\implies b=\frac{5\times (-4)+1\times (-6)}{5+1}\\</p><p>\implies b=\frac{-20-6}{6}\\</p><p>\implies b=\frac{-26}{6}\\</p><p>\implies b=\frac{-13}{3}$

Thus, the point of intersection is $P(0, \frac{-13}{3})$

Answer 2

Let the line segment A(5, -6) and B(-1, -4) is divided at point P(0, y) by y-axis in ratio m:n

:. x = $\frac{mx2+nx1}{m+n}$ and y = $\frac{my2+ny1}{m+n}$

Here, (x, y) = (0, y); (x1, y1) = (5, -6) and (x2, y2) = (-1, -4)

So , 0 = $\frac{m(-1)+n(5)}{m+n}$

=> 0 = -m + 5n

=> m= 5n

=> $\frac{m}{n}$ = $\frac{5}{1}$

=> m:n = 5:1

Hence, the ratio is 5:1 and the division is internal.Now,

y = $\frac{my2+ny1}{m+n}$

=> y = $\frac{5(-4)+1(-6)}{5+1}$

=> y = $\frac{-20-6}{6}$

=> y = $\frac{-26}{6}$

=> y = $\frac{-13}{3}$

Hence, the coordinates of the point of division is (0, -13/3).

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