Question

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

Answer 1

Step-by-step explanation:

Let the point be A(5,−6), B(−1,−4) and P(0,y)

Point P is on y−axis,hence its x co-ordinate is 0.So, it is of the form P(0,y)

Now, we have to find ratio.

Let the ratio be k:1

Hence m

1

=k,m

2

=1,x

1

=5,y

1

=−6,x

2

=−1,y

2

=−4,x=0,y=0

Using sections formula x=

m

1

+m

2

m

1

x

2

+m

2

x

1

⇒0=

k+1

−k+5

∴k=5

Again y=

m

1

+m

2

m

1

y

2

+m

2

y

1

=

k+1

−4k−6

=

6

−20−6

for k=5

=

3

−13

Hence the coordiantes of point is P(0,

3

−13

)

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.

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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