Question

COORDINATE GEOMETRY
CLASS 10
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•Please answer the question in the picture.
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THANKS...​ ​​please answerP​

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 :

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,

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

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,

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

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

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