Question

A,B and P are the three non-collinear points on a plane. The distance between the point A and P is 2m
more than the distance between the points B and P. if the distance between points A and B is 10m and AB
is the longest side of the triangle ABC. Is ABC a right angled triangle or not, Justify your answer using
the discriminant of quadratic equation and also find the measure of AP and BP.​

Answer 1

AnswEr :

$\bf{\large{\green{\underline{\underline{\tt{Given\::}}}}}}$

A,B and P are the three non - collinear points on a plan. The distance between the point A and P is 2 m more than the distance between the points B and P. If the distance between points A and B is 10 m and AB is the longest side of the Δ. Is ABC a right angled triangle or not.

$\bf{\large{\red{\underline{\underline{\bf{To\:find\::}}}}}}$

Is ABC a right angled triangle or not. and the measure of AP and BP.

$\bf{\Large{\pink{\underline{\underline{\rm{Explanation\::}}}}}}$

Reference of image according to the question :

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\put(7.7,2.9){\large{A}}\put(7.7,1){\large{P}}\put(10.6,1){\large{B}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(7.3,2){\sf{\large{(R+2)}}}\put(9,0.7){\sf{\large{R m}}}\put(9.4,1.9){\sf{\large{10 m}}}\put(8.2,1){\line(0.1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}}$

A/q

$\bf{We\:have}\begin{cases}\sf{\triangle APB \:is\:a\:right\:angled\:triangle}\\ \sf{AB\:=\:10m}\\ \sf{PB\:=\:R m}\\ \sf{AP\:=\:(R+2)m}\end{cases}}$

$\bf{\Large{\blue{\underline{\underline{\sf{\blacksquare{Using\:Pythagoras\:theorem\::}}}}}}}$

$\mapsto\sf{(Hypotenuse)^{2}=(base)^{2} +(perpendicular)^{2} }\\\\\\\\\mapsto\tt{(AB)^{2} =(PB)^{2} +(AP)^{2} }\\\\\\\\\mapsto\tt{(10)^{2} =(R)^{2} +(R+2)^{2}}\\\\\\\\\mapsto\tt{100=R^{2} +R^{2} +2^{2} +2*R*2}\\\\\\\\\mapsto\tt{100=2R^{2} +4+4R}\\\\\\\\\mapsto\tt{2R^{2} +4R+4-100=0}\\\\\\\\\mapsto\tt{2R^{2} +4R-96=0}\\\\\\\\\mapsto\tt{2(R^{2} +2R-48)=0}\\\\\\\\\mapsto\tt{R^{2}+2R-48=\cancel{\dfrac{0}{2}} }\\\\\\\\\mapsto\tt{\green{R^{2} +2R-48=0}}$

$\bf{\Large{\blue{\underline{\underline{\sf{\blacksquare{Using\:discriminant\:formula\::}}}}}}}$

$\leadsto\sf{x=\dfrac{-b\pm\sqrt{b^{2} -4ac} }{2a} }}$

We have;

• a = 1
• b = 2
• c = -48

$\dashrightarrow\tt{x=\dfrac{-2\pm\sqrt{(2)^{2}-4*1*(-48) } }{2*1}}\\\\\\\\\dashrightarrow\tt{x=\dfrac{-2\pm\sqrt{4-4(-48)} }{2} }\\\\\\\\\dashrightarrow\tt{x=\dfrac{-2\pm\sqrt{4+192} }{2} }\\\\\\\\\dashrightarrow\tt{x=\dfrac{-2\pm\sqrt{196} }{2} }\\\\\\\\\dashrightarrow\tt{x=\dfrac{-2\pm14}{2} }\\\\\\\\\dashrightarrow\tt{x=\dfrac{-2+14}{2} \:\:\:oR\:\:\:\dfrac{-2-14}{2}} \\\\\\\\\dashrightarrow\tt{x=\cancel{\dfrac{12}{2}} \:\:\:oR\:\:\:\cancel{\dfrac{-16}{2} }}$

$\dashrightarrow\tt{\red{x=6\:\:\:\:\:oR\:\:\:\:\:\:-8}}$

We know that negative value isn't acceptable.

So, x = 6

Now,

The length of PB = 6 m.

The length of AP = (6+2)m = 8 m.

Δ ABC isn't a right angled Δ.

Answer 2

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