Integrate it plzzz...

1.	Integrate it plzzz...
Answer» ANSWER :— <P> 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»

ANSWER :—

<P>

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,

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