Question

If x is a positive integer such that the distance between the points P(x,
2) and Q(3.-6) is 10 units, then find x.​

Answer 1

Answer:

Required value of x is 9.

Step-by-step explanation:

Using distance formula which says if coordinates of two points are ( a, b ) and ( c, d ) then distance x is given by :

• x = $\sqrt{(a-c)^2+(b-d)^2}$

Here,

points are ( x, 2 ) and ( 3 , - 6 ) and distance between them is 10 units :

$\implies\sqrt{(x-3)^2+(2+6)^2}=10$

= > ( x - 3 )^2 = 100 - 8^2

= > ( x - 3 )^2 = 100 - 64 = 36 = ( ± 6 )^2

= > x - 3 = 6 Or x - 3 = - 6

= > x = 6 + 3 Or x = - 6 + 3

= > x = 9 Or x = - 3

Since x is a positive integer, x ≠ - 3,

hence, x = 9

Required value of x is 9.

Answer 2

$\Large\underline\mathfrak{Question}$

• If x is a positive integer such that the distance between the points P(x,2) and Q(3.-6) is 10 units, then find x.

$\bf\red\bigstar\underline\textbf{ \blue{Extra}\:Brainly\: \pink{Knowledge}}\red\bigstar$ :-

✰) To find the distance between two points say P (x₁,y₁) and Q(x₂,y₂) is :

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

✰) To find the distance of a point say P (x,y) from origin is :

$\pink{\bf\sqrt {{x}^{2}+{y}^{2}}}$

✰) Coordinates of the point P (x,y) which device the line segment joining the points A (x₁,y₁) and B (x₂,y₂)

internally in the ratio m₁ : m₂ is :

$\red{\sf\dfrac {m_1 x_2 + m_2 x_1}{m_1+m_2},,,\dfrac {m_1 y_2 + m_2 y_1}{m_1+m_2}}$

✰) The midpoint of the line segment joining the points P (x₁,y₁) and Q(x₂,y₂) is : ---

$\purple{\tt\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}}$

✰) The area of the triangle formed by the points (x₁,y₁)(x₂,y₂) and (x_3,y_3) is the numerical value of the expression:----

$\green{\tt\dfrac{1}{2}[x_1(y_2-y_3)+x_2 (y_3-y_1)+x_3(y_1-y_2)]}$

$\rule{200}{4}$

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\:Answer}}}}}}}}}}$

$\begin{lgathered}\tt {\pink{Given}}\begin{cases} \sf{\green{P = (x, 2)}}\\ \sf{\blue{Q = (3, -6)}}\\ \sf{\orange{D = 10 units.}}\end{cases}\end{lgathered} \:$

$\large\red{\texttt{Putting values we get,}}$

$\purple\longmapsto\tt \: \sqrt{(3 - x)^{2} + ( - 6 - 2)^{2}} = 10 \\ \\ \bf \: squaring \: both \: sides \: we \: get \\ \\ \purple\longmapsto\tt \: (3 - x)^{2} + ( - 6 - 2)^{2} = 10^{2} \\ \\ \purple\longmapsto\tt \: (3 - x)^{2} = 100 - 64 \\ \\ \purple\longmapsto\tt \: (3 - x)^{2} =36 \\ \\ \bf \: square - root \: both \: sides \: now \\ \\ \purple\longmapsto\tt \: (3 - x) = \pm6 \\ \\ \purple\longmapsto\tt \: x =3 + 6 = 9 \\ \\ \: \: \huge\sf \: or \\ \\ \purple\longmapsto\tt \: x = 3 - 6 = ( - 3)$

$\large\red{\boxed{\tt\blue{x}\purple{=} \green {9} \orange,\pink{ - 3}}}$

$\rule{200}{4}$

$\red{\textbf{But it is Given that x is a positive integer,}}\\\sf\green{So,}\\\pink{\large\boxed{\boxed{\bold{x= 9}}}}$