Question

Find the ratio in which the line segment joining the points ( -3,10) and (6,-8) is divided by (-1,6)​

Answer 1

Answer:

Using the section formula, if a point (x,y) divides the line joining the points (x

1

,y

1

) and (x

2

,y

2

) in the ratio m:n, then

(x,y)=(

m+n

mx

2

+nx

1

,

m+n

my

2

+ny

1

)

Let the point P(−1,6) divides the line joining A(−3,10) and (6,−8) in the ratio k:1

Then, by section formula,

the coordinates of P are (

k+1

6k−3

,

k+1

−8k+10

)

⇒(−1,6)=(

k+1

6k−3

,

k+1

−8k+10

)

∴

k+1

6k−3

=−1

⇒6k−3=−k−1

⇒6k+k=3−1

⇒7k=2

∴ k=

7

2

Hence, the point P divides AB in the ratio 2:7

Answer 2

Answer:

• The ratio is 2 : 7.

Step-by-step explanation:

Given,

• A(-3, 10) and B(6, -8) is divided by (-1, 6).

To Find,

• The ratio in which it is divided.

Solution,

Let (-1, 6) divide AB internally in the ratio m1 : m2. Using the Section Formula, We get;

$: \implies ( x,y) = [ \frac{ m_{1}{x} _{2} + {m} _{2} {x} _{1}}{{m} _{1} + {m} _{2}}, \frac{{m} _{1}{y} _{2} +{m} _{2}{y} _{1} }{{m} _{1} + {m} _{2}} ] \\ \\ : \implies ( - 1,6) = [ \frac{ 6m_{1} - 3 {m} _{2}}{{m} _{1} + {m} _{2}}, \frac{ - 8{m} _{1}+10{m} _{2}}{{m} _{1} + {m} _{2}} ]$

So,

$: \implies - 1,= \frac{ 6m_{1} - 3 {m} _{2}}{{m} _{1} + {m} _{2}} \: \: and \: \: 6= \frac{ - 8{m} _{1}+10{m} _{2}}{{m} _{1} + {m} _{2}}$

Now,

$: \implies - 1= \frac{ 6m_{1} - 3 {m} _{2}}{{m} _{1} + {m} _{2}} \\ \\ : \implies - {m} _{1} - {m} _{2} = 6m_{1} - 3 {m} _{2} \\ \\ : \implies 2{m} _{2} = 7m_{1} \\ \\: \implies \color{red} \boxed{ m_{1} : {m} _{2} = 2 : 7}$

Required Answer,

• The ratio is 2 : 7.