Find the ratio in which the point P (3/4, 5/12) divides the line segment joining the points A(1/2, 3/2) and B(2, -5) example

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3. Find Ratio of line joining AB and is divided by P example ( Enter your problem )

( Enter your problem )

Find the ratio in which the point P (3/4, 5/12) divides the line segment joining the points A(1/2, 3/2) and B(2, -5)
Point that divides the line joining A(-4, 1) and B(17, 10) in the ratio 1 : 2
In what ratio does the x-axis divide the join of A(2, -3) and B (5, 6)
Find the ratio in which the point P(x,2) divides the line segment joining the points B(4,-3) and A(12,5)? Also find the value of x

Other related methods

2. Points are Collinear or Triangle or Quadrilateral form
(Previous method)

2. Point that divides the line joining A(-4, 1) and B(17, 10) in the ratio 1 : 2
(Next example)

1. Find the ratio in which the point P (3/4, 5/12) divides the line segment joining the points A(1/2, 3/2) and B(2, -5)

1. Find the ratio in which the point $P(3/4,5/12)$ divides the line segment joining the points $A(1/2,3/2)$ and $B(2,-5)$

Solution:
Method-1 : considering the ratio $m:n$

Suppose

$P(3/4,5/12)$ divides the line joining

$A(1/2,3/2)$ and

$B(2,-5)$ in the ratio

$m:n$

Using section formula

$P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))$

$:.P(3/4,5/12)=((m(2)+n(1/2))/(m+n),(m(-5)+n(3/2))/(m+n))$

$:.(m(2)+n(1/2))/(m+n)=3/4$ and

$(m(-5)+n(3/2))/(m+n)=5/12$

Now, solving first

$:. (m(2)+n(1/2))/(m+n)=3/4$

$:. 4(2m+(n)/(2))=3(m+n)$

$:. 8m+2n=3m+3n$

$:. 8m-3m=3n-2n$

$:. 5m=n$

$:. m/n=(1)/(5)$

$:.$ The point

$P(3/4,5/12)$ divides the line joining

$A(1/2,3/2)$ and

$B(2,-5)$ in the ratio

$1:5$

Method-2 : considering the ratio $k:1$

Suppose

$P(3/4,5/12)$ divides the line joining

$A(1/2,3/2)$ and

$B(2,-5)$ in the ratio

$k:1$

Using section formula

$P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))$

$:.P(3/4,5/12)=((k(2)+(1/2))/(k+1),(k(-5)+(3/2))/(k+1))$

$:.(k(2)+(1/2))/(k+1)=3/4$ and

$(k(-5)+(3/2))/(k+1)=5/12$

Now, solving first

$:. (k(2)+(1/2))/(k+1)=3/4$

$:. 4(2k+1/2)=3(k+1)$

$:. 8k+2=3k+3$

$:. 8k-3k=3-2$

$:. 5k=1$

$:. k=(1)/(5)$

$:.$ The point

$P(3/4,5/12)$ divides the line joining

$A(1/2,3/2)$ and

$B(2,-5)$ in the ratio

$1:5$

2. Find the ratio in which the point $P(-1,6)$ divides the line segment joining the points $A(3,10)$ and $B(6,-8)$

Solution:
Method-1 : considering the ratio $m:n$

Suppose

$P(-1,6)$ divides the line joining

$A(3,10)$ and

$B(6,-8)$ in the ratio

$m:n$

Using section formula

$P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))$

$:.P(-1,6)=((m(6)+n(3))/(m+n),(m(-8)+n(10))/(m+n))$

$:.(m(6)+n(3))/(m+n)=-1$ and

$(m(-8)+n(10))/(m+n)=6$

Now, solving first

$:. (m(6)+n(3))/(m+n)=-1$

$:. 6m+3n=-1(m+n)$

$:. 6m+3n=-m-n$

$:. 6m+m=-n-3n$

$:. 7m=-4n$

$:. m/n=(-4)/(7)$

As the ratio is negative, the point

$P(-1,6)$ divides the line joining

$A(3,10)$ and

$B(6,-8)$ externally in the ratio

$4:7$

Method-2 : considering the ratio $k:1$

Suppose

$P(-1,6)$ divides the line joining

$A(3,10)$ and

$B(6,-8)$ in the ratio

$k:1$

Using section formula

$P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))$

$:.P(-1,6)=((k(6)+(3))/(k+1),(k(-8)+(10))/(k+1))$

$:.(k(6)+(3))/(k+1)=-1$ and

$(k(-8)+(10))/(k+1)=6$

Now, solving first

$:. (k(6)+(3))/(k+1)=-1$

$:. 6k+3=-1(k+1)$

$:. 6k+3=-k-1$

$:. 6k+k=-1-3$

$:. 7k=-4$

$:. k=(-4)/(7)$

As the ratio is negative, the point

$P(-1,6)$ divides the line joining

$A(3,10)$ and

$B(6,-8)$ externally in the ratio

$4:7$

3. Find the ratio in which the point $P(-2,3)$ divides the line segment joining the points $A(-3,5)$ and $B(4,-9)$

Solution:
Method-1 : considering the ratio $m:n$

Suppose

$P(-2,3)$ divides the line joining

$A(-3,5)$ and

$B(4,-9)$ in the ratio

$m:n$

Using section formula

$P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))$

$:.P(-2,3)=((m(4)+n(-3))/(m+n),(m(-9)+n(5))/(m+n))$

$:.(m(4)+n(-3))/(m+n)=-2$ and

$(m(-9)+n(5))/(m+n)=3$

Now, solving first

$:. (m(4)+n(-3))/(m+n)=-2$

$:. 4m-3n=-2(m+n)$

$:. 4m-3n=-2m-2n$

$:. 4m+2m=-2n+3n$

$:. 6m=n$

$:. m/n=(1)/(6)$

$:.$ The point

$P(-2,3)$ divides the line joining

$A(-3,5)$ and

$B(4,-9)$ in the ratio

$1:6$

Method-2 : considering the ratio $k:1$

Suppose

$P(-2,3)$ divides the line joining

$A(-3,5)$ and

$B(4,-9)$ in the ratio

$k:1$

Using section formula

$P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))$

$:.P(-2,3)=((k(4)+(-3))/(k+1),(k(-9)+(5))/(k+1))$

$:.(k(4)+(-3))/(k+1)=-2$ and

$(k(-9)+(5))/(k+1)=3$

Now, solving first

$:. (k(4)+(-3))/(k+1)=-2$

$:. 4k-3=-2(k+1)$

$:. 4k-3=-2k-2$

$:. 4k+2k=-2+3$

$:. 6k=1$

$:. k=(1)/(6)$

$:.$ The point

$P(-2,3)$ divides the line joining

$A(-3,5)$ and

$B(4,-9)$ in the ratio

$1:6$

4. Find the ratio in which the point $P(3,10)$ divides the line segment joining the points $A(5,12)$ and $B(2,9)$

Solution:
Method-1 : considering the ratio $m:n$

Suppose

$P(3,10)$ divides the line joining

$A(5,12)$ and

$B(2,9)$ in the ratio

$m:n$

Using section formula

$P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))$

$:.P(3,10)=((m(2)+n(5))/(m+n),(m(9)+n(12))/(m+n))$

$:.(m(2)+n(5))/(m+n)=3$ and

$(m(9)+n(12))/(m+n)=10$

Now, solving first

$:. (m(2)+n(5))/(m+n)=3$

$:. 2m+5n=3(m+n)$

$:. 2m+5n=3m+3n$

$:. 2m-3m=3n-5n$

$:. -m=-2n$

$:. m=2n$

$:. m/n=(2)/(1)$

$:.$ The point

$P(3,10)$ divides the line joining

$A(5,12)$ and

$B(2,9)$ in the ratio

$2:1$

Method-2 : considering the ratio $k:1$

Suppose

$P(3,10)$ divides the line joining

$A(5,12)$ and

$B(2,9)$ in the ratio

$k:1$

Using section formula

$P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))$

$:.P(3,10)=((k(2)+(5))/(k+1),(k(9)+(12))/(k+1))$

$:.(k(2)+(5))/(k+1)=3$ and

$(k(9)+(12))/(k+1)=10$

Now, solving first

$:. (k(2)+(5))/(k+1)=3$

$:. 2k+5=3(k+1)$

$:. 2k+5=3k+3$

$:. 2k-3k=3-5$

$:. -k=-2$

$:. k=2$

$:. k=(2)/(1)$

$:.$ The point

$P(3,10)$ divides the line joining

$A(5,12)$ and

$B(2,9)$ in the ratio

$2:1$

5. Find the ratio in which the point $P(6,17)$ divides the line segment joining the points $A(1,-3)$ and $B(3,5)$

Solution:
Method-1 : considering the ratio $m:n$

Suppose

$P(6,17)$ divides the line joining

$A(1,-3)$ and

$B(3,5)$ in the ratio

$m:n$

Using section formula

$P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))$

$:.P(6,17)=((m(3)+n(1))/(m+n),(m(5)+n(-3))/(m+n))$

$:.(m(3)+n(1))/(m+n)=6$ and

$(m(5)+n(-3))/(m+n)=17$

Now, solving first

$:. (m(3)+n(1))/(m+n)=6$

$:. 3m+n=6(m+n)$

$:. 3m+n=6m+6n$

$:. 3m-6m=6n-n$

$:. -3m=5n$

$:. 3m=-5n$

$:. m/n=(-5)/(3)$

As the ratio is negative, the point

$P(6,17)$ divides the line joining

$A(1,-3)$ and

$B(3,5)$ externally in the ratio

$5:3$

Method-2 : considering the ratio $k:1$

Suppose

$P(6,17)$ divides the line joining

$A(1,-3)$ and

$B(3,5)$ in the ratio

$k:1$

Using section formula

$P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))$

$:.P(6,17)=((k(3)+(1))/(k+1),(k(5)+(-3))/(k+1))$

$:.(k(3)+(1))/(k+1)=6$ and

$(k(5)+(-3))/(k+1)=17$

Now, solving first

$:. (k(3)+(1))/(k+1)=6$

$:. 3k+1=6(k+1)$

$:. 3k+1=6k+6$

$:. 3k-6k=6-1$

$:. -3k=5$

$:. 3k=-5$

$:. k=(-5)/(3)$

As the ratio is negative, the point

$P(6,17)$ divides the line joining

$A(1,-3)$ and

$B(3,5)$ externally in the ratio

$5:3$

6. Find the ratio in which the point $P(12,23)$ divides the line segment joining the points $A(2,8)$ and $B(6,14)$

Solution:
Method-1 : considering the ratio $m:n$

Suppose

$P(12,23)$ divides the line joining

$A(2,8)$ and

$B(6,14)$ in the ratio

$m:n$

Using section formula

$P(x,y)=((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))$

$:.P(12,23)=((m(6)+n(2))/(m+n),(m(14)+n(8))/(m+n))$

$:.(m(6)+n(2))/(m+n)=12$ and

$(m(14)+n(8))/(m+n)=23$

Now, solving first

$:. (m(6)+n(2))/(m+n)=12$

$:. 6m+2n=12(m+n)$

$:. 6m+2n=12m+12n$

$:. 6m-12m=12n-2n$

$:. -6m=10n$

$:. 6m=-10n$

$:. m/n=(-10)/(6)$

$:. m/n=(-5)/(3)$

As the ratio is negative, the point

$P(12,23)$ divides the line joining

$A(2,8)$ and

$B(6,14)$ externally in the ratio

$5:3$

Method-2 : considering the ratio $k:1$

Suppose

$P(12,23)$ divides the line joining

$A(2,8)$ and

$B(6,14)$ in the ratio

$k:1$

Using section formula

$P(x,y)=((kx_2+x_1)/(k+1),(ky_2+y_1)/(k+1))$

$:.P(12,23)=((k(6)+(2))/(k+1),(k(14)+(8))/(k+1))$

$:.(k(6)+(2))/(k+1)=12$ and

$(k(14)+(8))/(k+1)=23$

Now, solving first

$:. (k(6)+(2))/(k+1)=12$

$:. 6k+2=12(k+1)$

$:. 6k+2=12k+12$

$:. 6k-12k=12-2$

$:. -6k=10$

$:. 6k=-10$

$:. k=(-10)/(6)$

$:. k=(-5)/(3)$

As the ratio is negative, the point

$P(12,23)$ divides the line joining

$A(2,8)$ and

$B(6,14)$ externally in the ratio

$5:3$

This material is intended as a summary. Use your textbook for detail explanation.
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2. Points are Collinear or Triangle or Quadrilateral form
(Previous method)

2. Point that divides the line joining A(-4, 1) and B(17, 10) in the ratio 1 : 2
(Next example)

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