Question

Please help me out with this Q6!!

Answer 1

Answer:

$3x + 4y = 7$

A(1,2) and B(-2,1)

Let ratio is k : 1

$x1 = 1 \\ x2 = - 2 \\ y1 = 2 \\ y2 = 1 \\ m = k \\ n = 1$

By Section Formula

P(x,y)

$=( \frac{mx2 + nx1}{m + n} \: \frac{my2 + ny1}{m + n} ) \\ \\ = ( \frac{k( - 2) + 1(1)}{k + 1} \: \frac{k(1) + 1(2)}{k + 1} ) \\ \\ = ( \frac{ - 2k + 1}{k + 1} \: \frac{k + 2}{k + 1})$

By comparing coordinates

$x = \frac{ - 2k + 1}{k + 1} \\ \\ y = \frac{k + 2}{k + 1}$

Put values of x and y in the given equation.

$3x + 4y = 7 \\ 3( \frac{ - 2k + 1}{k + 1} ) + 4( \frac{k + 2}{k + 1} ) = 7 \\ \\ \frac{ - 6k + 3}{k + 1} + \frac{4k + 8}{k + 1} = 7 \\ \\ \frac{ - 6k + 3 + 4k + 8}{k + 1} = 7 \\ \\ 11 - 2k = 7k + 7 \\ 4 = 9k \\ \frac{k}{1} = \frac{4}{9}$

Therefore, (c) 4 : 9

HOPE THIS ANSWER WILL HELP YOU..........

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# jasleenlehri13.........☺✌

Answer 2

Answer:

please mark me as brainliest

Step-by-step explanation:

3x+4y=7

A(1,2) and B(-2,1)

Let ratio is k : 1

\begin{lgathered}x1 = 1 \\ x2 = - 2 \\ y1 = 2 \\ y2 = 1 \\ m = k \\ n = 1\end{lgathered}

x1=1

x2=−2

y1=2

y2=1

m=k

n=1

By Section Formula

P(x,y)

\begin{lgathered}=( \frac{mx2 + nx1}{m + n} \: \frac{my2 + ny1}{m + n} ) \\ \\ = ( \frac{k( - 2) + 1(1)}{k + 1} \: \frac{k(1) + 1(2)}{k + 1} ) \\ \\ = ( \frac{ - 2k + 1}{k + 1} \: \frac{k + 2}{k + 1})\end{lgathered}

=(

m+n

mx2+nx1

m+n

my2+ny1

)

=(

k+1

k(−2)+1(1)

k+1

k(1)+1(2)

)

=(

k+1

−2k+1

k+1

k+2

)

By comparing coordinates

\begin{lgathered}x = \frac{ - 2k + 1}{k + 1} \\ \\ y = \frac{k + 2}{k + 1}\end{lgathered}

x=

k+1

−2k+1

y=

k+1

k+2

Put values of x and y in the given equation.

\begin{lgathered}3x + 4y = 7 \\ 3( \frac{ - 2k + 1}{k + 1} ) + 4( \frac{k + 2}{k + 1} ) = 7 \\ \\ \frac{ - 6k + 3}{k + 1} + \frac{4k + 8}{k + 1} = 7 \\ \\ \frac{ - 6k + 3 + 4k + 8}{k + 1} = 7 \\ \\ 11 - 2k = 7k + 7 \\ 4 = 9k \\ \frac{k}{1} = \frac{4}{9}\end{lgathered}

3x+4y=7

3(

k+1

−2k+1

)+4(

k+1

k+2

)=7

k+1

−6k+3

+

k+1

4k+8

=7

k+1

−6k+3+4k+8

=7

11−2k=7k+7

4=9k

1

k

=

9

4

Therefore, (c) 4 : 9