3. In what ratio does the x-axis divide the join of A(2, -3) and B (5, 6)
1. In what ratio does the x-axis divide the join of `A(2,-3)` and `B(5,6)`? Also find the coordinates of the point of intersection.
Solution:
Method-1 : considering the ratio `m:n`
Let x-axis divides the line segment joining the points `A(2,-3)` and `B(5,6)` in the ratio `m:n` at the point `P`
The given points are `A(2,-3),B(5,6)`
`:. x_1=2,y_1=-3,x_2=5,y_2=6`
Using section formula
`P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))`
`:.P(x,y)=((m(5)+n(2))/(m+n),(m(6)+n(-3))/(m+n))`
On the x-axis the y-coordinate of every point is 0, i.e. `y=0`
`:. (m(6)+n(-3))/(m+n)=0`
`:. 6m-3n=0`
`:. 6m=3n`
`:. m/n=(3)/(6)`
`:. m/n=(1)/(2)`
:. The point divides the line joining `A(2,-3)` and `B(5,6)` in the ratio `1:2`
Putting `m=1,n=2`, we get the coordinates of point P
`x=(mx_2+nx_1)/(m+n)`
`=(1*5+2*2)/(1+2)`
`=(5+4)/(3)`
`=(9)/(3)`
`=3`
Hence, the point of intersection of AB and the x-axis is `P(3,0)`
Method-2 : considering the ratio `k:1`
Let x-axis divides the line segment joining the points `A(2,-3)` and `B(5,6)` in the ratio `k:1` at the point `P`
The given points are `A(2,-3),B(5,6)`
`:. x_1=2,y_1=-3,x_2=5,y_2=6`
Using section formula
`P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))`
`:.P(x,y)=((k(5)+(2))/(k+1),(k(6)+(-3))/(k+1))`
On the x-axis the y-coordinate of every point is 0, i.e. `y=0`
`:. (k(6)+(-3))/(k+1)=0`
`:. 6k-3=0`
`:. 6k=3`
`:. k=(3)/(6)`
`:. k=(1)/(2)`
:. The point divides the line joining `A(2,-3)` and `B(5,6)` in the ratio `1:2`
Putting `k=1/2`, we get the coordinates of point P
`x=(kx_2+x_1)/(k+1)`
`=(1/2*5+2)/(1/2+1)`
`=(1*5+2*2)/(1+2)`
`=(5+4)/(3)`
`=(9)/(3)`
`=3`
Hence, the point of intersection of AB and the x-axis is `P(3,0)`
2. In what ratio does the x-axis divide the join of `A(1,2)` and `B(2,3)`? Also find the coordinates of the point of intersection.
Solution:
Method-1 : considering the ratio `m:n`
Let x-axis divides the line segment joining the points `A(1,2)` and `B(2,3)` in the ratio `m:n` at the point `P`
The given points are `A(1,2),B(2,3)`
`:. x_1=1,y_1=2,x_2=2,y_2=3`
Using section formula
`P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))`
`:.P(x,y)=((m(2)+n(1))/(m+n),(m(3)+n(2))/(m+n))`
On the x-axis the y-coordinate of every point is 0, i.e. `y=0`
`:. (m(3)+n(2))/(m+n)=0`
`:. 3m+2n=0`
`:. 3m=-2n`
`:. m/n=(-2)/(3)`
As the ratio is negative, the point divides the line joining `A(1,2)` and `B(2,3)` externally in the ratio `2:3`
Putting `m=-2,n=3`, we get the coordinates of point P
`x=(mx_2+nx_1)/(m+n)`
`=(-2*2+3*1)/(-2+3)`
`=(-4+3)/(1)`
`=(-1)/(1)`
`=-1`
Hence, the point of intersection of AB and the x-axis is `P(-1,0)`
Method-2 : considering the ratio `k:1`
Let x-axis divides the line segment joining the points `A(1,2)` and `B(2,3)` in the ratio `k:1` at the point `P`
The given points are `A(1,2),B(2,3)`
`:. x_1=1,y_1=2,x_2=2,y_2=3`
Using section formula
`P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))`
`:.P(x,y)=((k(2)+(1))/(k+1),(k(3)+(2))/(k+1))`
On the x-axis the y-coordinate of every point is 0, i.e. `y=0`
`:. (k(3)+(2))/(k+1)=0`
`:. 3k+2=0`
`:. 3k=-2`
`:. k=(-2)/(3)`
As the ratio is negative, the point divides the line joining `A(1,2)` and `B(2,3)` externally in the ratio `2:3`
Putting `k=-2/3`, we get the coordinates of point P
`x=(kx_2+x_1)/(k+1)`
`=(-2/3*2+1)/(-2/3+1)`
`=(-2*2+1*3)/(-2+3)`
`=(-4+3)/(1)`
`=(-1)/(1)`
`=-1`
Hence, the point of intersection of AB and the x-axis is `P(-1,0)`
3. In what ratio does the y-axis divide the join of `A(5,-6)` and `B(-1,-4)`? Also find the coordinates of the point of intersection.
Solution:
Method-1 : considering the ratio `m:n`
Let y-axis divides the line segment joining the points `A(5,-6)` and `B(-1,-4)` in the ratio `m:n`
The given points are `A(5,-6),B(-1,-4)`
`:. x_1=5,y_1=-6,x_2=-1,y_2=-4`
Using section formula
`P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))`
`:.P(x,y)=((m(-1)+n(5))/(m+n),(m(-4)+n(-6))/(m+n))`
On the y-axis the x-coordinate of every point is 0, i.e. `x=0`
`:. (m(-1)+n(5))/(m+n)=0`
`:. -m+5n=0`
`:. -m=-5n`
`:. m=5n`
`:. m/n=(5)/(1)`
The point divides the line joining `A(5,-6)` and `B(-1,-4)` in the ratio `5:1`
Putting `m=5,n=1`, we get the coordinates of point P
`y=(my_2+ny_1)/(m+n)`
`=(5*-4+1*-6)/(5+1)`
`=(-20-6)/(6)`
`=(-26)/(6)`
`=-13/3`
Hence, the point of intersection of AB and the y-axis is `P(0,-13/3)`
Method-2 : considering the ratio `k:1`
Let y-axis divides the line segment joining the points `A(5,-6)` and `B(-1,-4)` in the ratio `k:1` at the point `P`
The given points are `A(5,-6),B(-1,-4)`
`:. x_1=5,y_1=-6,x_2=-1,y_2=-4`
Using section formula
`P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))`
`:.P(x,y)=((k(-1)+(5))/(k+1),(k(-4)+(-6))/(k+1))`
On the y-axis the x-coordinate of every point is 0, i.e. `x=0`
`:. (k(-1)+(5))/(k+1)=0`
`:. -k+5=0`
`:. -k=-5`
`:. k=5`
`:. k=(5)/(1)`
The point divides the line joining `A(5,-6)` and `B(-1,-4)` in the ratio `5:1`
Putting `k=5`, we get the coordinates of point P
`y=(ky_2+y_1)/(k+1)`
`=(5*-4-6)/(5+1)`
`=(-20-6)/(6)`
`=(-26)/(6)`
`=-13/3`
Hence, the point of intersection of AB and the y-axis is `P(0,-13/3)`
4. In what ratio does the y-axis divide the join of `A(-2,1)` and `B(4,5)`? Also find the coordinates of the point of intersection.
Solution:
Method-1 : considering the ratio `m:n`
Let y-axis divides the line segment joining the points `A(-2,1)` and `B(4,5)` in the ratio `m:n`
The given points are `A(-2,1),B(4,5)`
`:. x_1=-2,y_1=1,x_2=4,y_2=5`
Using section formula
`P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))`
`:.P(x,y)=((m(4)+n(-2))/(m+n),(m(5)+n(1))/(m+n))`
On the y-axis the x-coordinate of every point is 0, i.e. `x=0`
`:. (m(4)+n(-2))/(m+n)=0`
`:. 4m-2n=0`
`:. 4m=2n`
`:. m/n=(2)/(4)`
`:. m/n=(1)/(2)`
The point divides the line joining `A(-2,1)` and `B(4,5)` in the ratio `1:2`
Putting `m=1,n=2`, we get the coordinates of point P
`y=(my_2+ny_1)/(m+n)`
`=(1*5+2*1)/(1+2)`
`=(5+2)/(3)`
`=(7)/(3)`
Hence, the point of intersection of AB and the y-axis is `P(0,7/3)`
Method-2 : considering the ratio `k:1`
Let y-axis divides the line segment joining the points `A(-2,1)` and `B(4,5)` in the ratio `k:1` at the point `P`
The given points are `A(-2,1),B(4,5)`
`:. x_1=-2,y_1=1,x_2=4,y_2=5`
Using section formula
`P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))`
`:.P(x,y)=((k(4)+(-2))/(k+1),(k(5)+(1))/(k+1))`
On the y-axis the x-coordinate of every point is 0, i.e. `x=0`
`:. (k(4)+(-2))/(k+1)=0`
`:. 4k-2=0`
`:. 4k=2`
`:. k=(2)/(4)`
`:. k=(1)/(2)`
The point divides the line joining `A(-2,1)` and `B(4,5)` in the ratio `1:2`
Putting `k=1/2`, we get the coordinates of point P
`y=(ky_2+y_1)/(k+1)`
`=(1/2*5+1)/(1/2+1)`
`=(1*5+1*2)/(1+2)`
`=(5+2)/(3)`
`=(7)/(3)`
Hence, the point of intersection of AB and the y-axis is `P(0,7/3)`
5. In what ratio does the x-axis divide the join of `A(2,1)` and `B(7,6)`? Also find the coordinates of the point of intersection.
Solution:
Method-1 : considering the ratio `m:n`
Let x-axis divides the line segment joining the points `A(2,1)` and `B(7,6)` in the ratio `m:n` at the point `P`
The given points are `A(2,1),B(7,6)`
`:. x_1=2,y_1=1,x_2=7,y_2=6`
Using section formula
`P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))`
`:.P(x,y)=((m(7)+n(2))/(m+n),(m(6)+n(1))/(m+n))`
On the x-axis the y-coordinate of every point is 0, i.e. `y=0`
`:. (m(6)+n(1))/(m+n)=0`
`:. 6m+n=0`
`:. 6m=-n`
`:. m/n=(-1)/(6)`
As the ratio is negative, the point divides the line joining `A(2,1)` and `B(7,6)` externally in the ratio `1:6`
Putting `m=-1,n=6`, we get the coordinates of point P
`x=(mx_2+nx_1)/(m+n)`
`=(-1*7+6*2)/(-1+6)`
`=(-7+12)/(5)`
`=(5)/(5)`
`=1`
Hence, the point of intersection of AB and the x-axis is `P(1,0)`
Method-2 : considering the ratio `k:1`
Let x-axis divides the line segment joining the points `A(2,1)` and `B(7,6)` in the ratio `k:1` at the point `P`
The given points are `A(2,1),B(7,6)`
`:. x_1=2,y_1=1,x_2=7,y_2=6`
Using section formula
`P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))`
`:.P(x,y)=((k(7)+(2))/(k+1),(k(6)+(1))/(k+1))`
On the x-axis the y-coordinate of every point is 0, i.e. `y=0`
`:. (k(6)+(1))/(k+1)=0`
`:. 6k+1=0`
`:. 6k=-1`
`:. k=(-1)/(6)`
As the ratio is negative, the point divides the line joining `A(2,1)` and `B(7,6)` externally in the ratio `1:6`
Putting `k=-1/6`, we get the coordinates of point P
`x=(kx_2+x_1)/(k+1)`
`=(-1/6*7+2)/(-1/6+1)`
`=(-1*7+2*6)/(-1+6)`
`=(-7+12)/(5)`
`=(5)/(5)`
`=1`
Hence, the point of intersection of AB and the x-axis is `P(1,0)`
6. In what ratio does the y-axis divide the join of `A(2,-4)` and `B(-3,6)`? Also find the coordinates of the point of intersection.
Solution:
Method-1 : considering the ratio `m:n`
Let y-axis divides the line segment joining the points `A(2,-4)` and `B(-3,6)` in the ratio `m:n`
The given points are `A(2,-4),B(-3,6)`
`:. x_1=2,y_1=-4,x_2=-3,y_2=6`
Using section formula
`P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))`
`:.P(x,y)=((m(-3)+n(2))/(m+n),(m(6)+n(-4))/(m+n))`
On the y-axis the x-coordinate of every point is 0, i.e. `x=0`
`:. (m(-3)+n(2))/(m+n)=0`
`:. -3m+2n=0`
`:. -3m=-2n`
`:. 3m=2n`
`:. m/n=(2)/(3)`
The point divides the line joining `A(2,-4)` and `B(-3,6)` in the ratio `2:3`
Putting `m=2,n=3`, we get the coordinates of point P
`y=(my_2+ny_1)/(m+n)`
`=(2*6+3*-4)/(2+3)`
`=(12-12)/(5)`
`=(0)/(5)`
Hence, the point of intersection of AB and the y-axis is `P(0,0)`
Method-2 : considering the ratio `k:1`
Let y-axis divides the line segment joining the points `A(2,-4)` and `B(-3,6)` in the ratio `k:1` at the point `P`
The given points are `A(2,-4),B(-3,6)`
`:. x_1=2,y_1=-4,x_2=-3,y_2=6`
Using section formula
`P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))`
`:.P(x,y)=((k(-3)+(2))/(k+1),(k(6)+(-4))/(k+1))`
On the y-axis the x-coordinate of every point is 0, i.e. `x=0`
`:. (k(-3)+(2))/(k+1)=0`
`:. -3k+2=0`
`:. -3k=-2`
`:. 3k=2`
`:. k=(2)/(3)`
The point divides the line joining `A(2,-4)` and `B(-3,6)` in the ratio `2:3`
Putting `k=2/3`, we get the coordinates of point P
`y=(ky_2+y_1)/(k+1)`
`=(2/3*6-4)/(2/3+1)`
`=(2*6-4*3)/(2+3)`
`=(12-12)/(5)`
`=(0)/(5)`
Hence, the point of intersection of AB and the y-axis is `P(0,0)`
This material is intended as a summary. Use your textbook for detail explanation.
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