Question

integrate it plzzz... ​

Answer 1

AnswEr :—

Given that,

Points P(p,-2) and Q(5/3,q) trisect the line AB.

The given points are A(3,-4) and B(1,2)

Now,

P(p,-2) divides the line in the ratio 1:2.

$\setlength{\unitlength}{1cm}\thicklines\begin{picture}(10,6)\put(0,0){\line(1,0){9}}\put(0,0){\circle*{0.1}}\put(0,-0.5){\sf A\;(3,-4)}\put(3,0){\circle*{0.1}}\put(3,-0.5){\sf P\;(7/3,-2)}\put(6,0){\circle*{0.1}}\put(6,-0.5){\sf Q\;(5/3,0)}\put(9,0){\circle*{0.1}}\put(9,-0.5){\sf B\;(1,2)}\end{picture}$

By using Sector's formula,

$\star \ \boxed{\boxed{\sf Z(x,y) = \bigg(\dfrac{nx_2 + mx_1}{m + n},\dfrac{ny_2 + my_1}{m + n} \bigg)}}$

Thus,

$\sf P(p,-2) = \bigg(\dfrac{1(1) + 3(2)}{1+2},\dfrac{2(1) -4(2)}{1+2} \bigg)$

Therefore,

$\implies \sf p = \dfrac{1 + 6}{3} \\ \\ \implies \boxed{ \boxed{\sf p = \dfrac{7}{3} }}$

Also,

Q is the midpoint of line PB.

By using Midpoint formula,

$\star \ \boxed{\boxed{\sf Z(x,y) = \bigg( \dfrac{x_1 + x_2}{2},\dfrac{y_1 + y_2}{2} \bigg)}}$

Thus,

$\sf Q(\dfrac{5}{3},q) = \bigg(\dfrac{\frac{7}{3} + 1}{2},\dfrac{-2 + 2}{1+2} \bigg)$

Therefore,

$\implies \sf \: q = \dfrac{2 - 2}{2} \\ \\ \implies \boxed{ \boxed{ \sf q = 0}}$

• The values of p and q are 7/3 and 0 respectively

Answer 2

Answer:

Points P(p,-2) and Q(5/3,q) trisect the line AB.

The given points are A(3,-4) and B(1,2)

Now,

P(p,-2) divides the line in the ratio 1:2.

By using Sector's formula,

\star \ \boxed{\boxed{\sf Z(x,y) = \bigg(\dfrac{nx_2 + mx_1}{m + n},\dfrac{ny_2 + my_1}{m + n} \bigg)}}⋆

Z(x,y)=(

m+n

nx

2

+mx

1

,

m+n

ny

2

+my

1

)

Thus,

\sf P(p,-2) = \bigg(\dfrac{1(1) + 3(2)}{1+2},\dfrac{2(1) -4(2)}{1+2} \bigg)P(p,−2)=(

1+2

1(1)+3(2)

,

1+2

2(1)−4(2)

)

Therefore,

\begin{gathered} \implies \sf p = \dfrac{1 + 6}{3} \\ \\ \implies \boxed{ \boxed{\sf p = \dfrac{7}{3} }}\end{gathered}

⟹p=

3

1+6

⟹

p=

3

7

Also,

Q is the midpoint of line PB.

By using Midpoint formula,

\star \ \boxed{\boxed{\sf Z(x,y) = \bigg( \dfrac{x_1 + x_2}{2},\dfrac{y_1 + y_2}{2} \bigg)}}⋆

Z(x,y)=(

2

x

1

+x

2

,

2

y

1

+y

2

)

Thus,

\sf Q(\dfrac{5}{3},q) = \bigg(\dfrac{\frac{7}{3} + 1}{2},\dfrac{-2 + 2}{1+2} \bigg)Q(

3

5

,q)=(

2

3

7

+1

,

1+2

−2+2

)

Therefore,

\begin{gathered} \implies \sf \: q = \dfrac{2 - 2}{2} \\ \\ \implies \boxed{ \boxed{ \sf q = 0}}\end{gathered}

⟹q=

2

2−2

⟹

q=0

• The values of p and q are 7/3 and 0 respectively

Step-by-step explanation:

$thanks$