Question

The points A(4,7) , B(p,3) and C(7,3) are the vertices of a right triangle, right angled at B . find the value of p.

Answer 1

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Refer to attachment :)

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Answer 2

Answer:

The value of  can be 4 or 7

Step-by-step explanation:

The points A(4,7) , B(p,3) and C(7,3) are the vertices of a right triangle, right angled at B

So, AB = perpendicular

BC = Base

AC = Hypotenuse

 A=(4,7) , B=(p,3)

We will use distance formula to calculate the length of AB

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(x_1,y_1)=(4,7)$

$(x_2,y_2)=(p,3)$

Length of AB = $\sqrt{(p-4)^2+(7-3)^2}$

Length of AB = $\sqrt{p^2+16-8p+16}$

Length of AB = $\sqrt{p^2-8p+32}$

To Find Length of BC

$(x_1,y_1)=(p,3)$

$(x_2,y_2)=(7,3)$

Length of BC = $\sqrt{(7-p)^2+(3-3)^2}$

Length of BC = $\sqrt{49+p^2-14p}$

To Find Length of AC

$(x_1,y_1)=(4,7)$

$(x_2,y_2)=(7,3)$

Length of AC = $\sqrt{(7-4)^2+(3-7)^2}$

Length of AC = $\sqrt{9+16}$

Length of AC = 5

So, Hypotenuse = AC = 5

Base =  BC = $\sqrt{49+p^2-14p}$

Perpendicular = AB = $\sqrt{p^2-8p+32}$

$Hypotenuse^2=Perpendicular^2+Base^2$

$5^2=(\sqrt{p^2-8p+32})^2+(\sqrt{49+p^2-14p})^2$

$25=p^2-8p+32+49+p^2-14p$

$25=2p^2-22p+81$

$2p^2-22p+56=0$

$p^2-11p+28=0$

$p^2-7p-4p+28=0$

$p(p-7)-4(p-7)=0$

$(p-7)(p-4)=0$

So, p = 4,7

Hence the value of  can be 4 or 7