find the ratio in which the y axis divides the line segment joining the points (5,-6) and (-1,-4). Also find the point of intersection according to m1:m2 ratio. About the author Remi
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 = [tex]\frac{mx2+nx1}{m+n}[/tex] and y = [tex]\frac{my2+ny1}{m+n}[/tex] Here, (x, y) = (0, y); (x1, y1) = (5, -6) and (x2, y2) = (-1, -4) So , 0 = [tex]\frac{m(-1)+n(5)}{m+n}[/tex] => 0 = -m + 5n => m= 5n => [tex]\frac{m}{n}[/tex] = [tex]\frac{5}{1}[/tex] => m:n = 5:1 Hence, the ratio is 5:1 and the division is internal.Now, y = [tex]\frac{my2+ny1}{m+n}[/tex] => y = [tex]\frac{5(-4)+1(-6)}{5+1}[/tex] => y = [tex]\frac{-20-6}{6}[/tex] => y = [tex]\frac{-26}{6}[/tex] => y = [tex]\frac{-13}{3}[/tex] Hence, the coordinates of the point of division is (0, -13/3). [tex]\\\\\\[/tex] HOPE IT HELPS PLEASE MARK ME BRAINLIEST ☺️ Reply
x1=5;y1=-6 x2=-1;y2=-4 Therefore, mx2+nx1/m+n 0=m(-1)+n(5)/m+n 0=-1m+5n/m+n -1m+5n=0 m/n=5/1 5:1 Reply
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 = [tex]\frac{mx2+nx1}{m+n}[/tex] and y = [tex]\frac{my2+ny1}{m+n}[/tex]
Here, (x, y) = (0, y); (x1, y1) = (5, -6) and (x2, y2) = (-1, -4)
So , 0 = [tex]\frac{m(-1)+n(5)}{m+n}[/tex]
=> 0 = -m + 5n
=> m= 5n
=> [tex]\frac{m}{n}[/tex] = [tex]\frac{5}{1}[/tex]
=> m:n = 5:1
Hence, the ratio is 5:1 and the division is internal.Now,
y = [tex]\frac{my2+ny1}{m+n}[/tex]
=> y = [tex]\frac{5(-4)+1(-6)}{5+1}[/tex]
=> y = [tex]\frac{-20-6}{6}[/tex]
=> y = [tex]\frac{-26}{6}[/tex]
=> y = [tex]\frac{-13}{3}[/tex]
Hence, the coordinates of the point of division is (0, -13/3).
[tex]\\\\\\[/tex]
HOPE IT HELPS
PLEASE MARK ME BRAINLIEST ☺️
x1=5;y1=-6
x2=-1;y2=-4
Therefore,
mx2+nx1/m+n
0=m(-1)+n(5)/m+n
0=-1m+5n/m+n
-1m+5n=0
m/n=5/1
5:1