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if P(p, p^2) lies inside the triangle having sides along the lines \\ {2 x+3 y=1, x+2 y-3=0,6 y=5 x-1} then the value of p ?

Option: 1

p \epsilon (1 / 2,1) \cup(-3 / 2,-1)


Option: 2

p \epsilon (1 / 2,1) \cup(-2,-1)


Option: 3

p \epsilon (1 ,2) \cup(-3 / 2,-1)


Option: 4

None of these


Answers (1)

best_answer

 

 

Position of two points with respect to a line -

Position of two points with respect to a line

Two given points and lies on the same side of a line ax + by + c = 0 when   and points lie on the opposite side when  .

Note:

  1. The side of the line where origin lies is known as the origin side.

  2. A point (p, q) will lie on the origin side of the line ax + by + c = 0 if ap + bq + c and c will have the same sign.

  3. A point (p, q) will lie on the non-origin side of the line ax + by + c = 0, if ap + bq + c and c will have the opposite sign.

Position of a point which lies inside a triangle

Let P(x1 ,y1) be the point that lies inside the triangle.

The equations of sides of a triangle are

\\\mathrm{AB:a_1x+b_1y+c_1=0}\\\mathrm{BC:a_2x+b_2y+c_2=0}\\\mathrm{CA:a_3x+b_3y+c_3=0}

First find the coordinates of vertices of triangle ABC

Let \\A=\left ( x',y' \right ),\;\;B=\left ( x'',y'' \right )\;\;and\;\;C=\left ( x''',y''' \right )

And if coordinates of vertices of triangle ABC is given then find equation of sides of triangle ABC.

If point P lies inside the triangle, then P and A must be same side of BC, P and B must be same side of AC and P and C must be same side of AB, then.

\\\frac{a_2x_1+b_2y_1+c_2}{a_2x'+b_2y'+c_2}>0\\\\\frac{a_3x_1+b_3y_1+c_3}{a_3x''+b_3y''+c_3}>0\\\\\frac{a_1x_1+b_1y_1+c_1}{a_1x'''+b_1y'1'+c_3}>0

-

 

 

 

\\ 2 x+3 y=1...(i)\\ x+2 y=3...(ii)\\5 x-6y=1..(iii)\\ \text{equation (i)-2*equation(ii)}\\ Point A(-7,5)\\ \text{5*equation (ii)-equation(iii)}\\ point B(5/4,7/8)\\ \text{5*equation (i)-2*equation(iii)}\\ Point C(1/3,1/9)\\

\begin{array}{l}{\text { So } \mathrm{A}, \mathrm{B}, \mathrm{C} \text { be vertices of the triangle. }} \\ {\mathrm{A} \equiv(-7,5), \mathrm{B} \equiv(5 / 4,7 / 8)} \\ {\mathrm{C} \equiv(1 / 3,1 / 9) } \\ \end{array}

\begin{array}{l}{\text { If P lies in-side the } \triangle \mathrm{ABC} \text { , then sign of P will be the same as sign of a w.r.t. the line BC }} \\ {\Rightarrow \quad 5 p-6 p^{2}-1<0 \Rightarrow (-\infty,\frac{1}{3})U(\frac{1}{2},\infty)} \\ {\text { Similarly } 2 p+3 p^{2}-1>0 \Rightarrow (-\infty,-1)U(\frac{1}{3},\infty)\text { . }} \\ {\text { And, } \quad p+2 p^{2}-3<0\Rightarrow (\frac{-3}{2},1) \text { . }} \\ {\text { Solving, }(1),(2) \text { and }(3) \text { for } p \text { and then taking intersection, }} \\ {\text { We get } \quad p \epsilon (1 / 2,1) \cup(-3 / 2,-1) \text { . }}\end{array}

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seema garhwal

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