Question

If the distance between (2,-3) and (10,x) is
10 units then find x.​

Answer 1

Given :-

• Distance between (2, - 3) and (10, x) is 10 units

To find :-

• Find the value of x

Solution :-

Apply distance formula

→ AB = √(x2 - x1)² + (y2 - y1)²

Let

• A = (2, - 3)
• B = (10, x)
• AB = 10

→ AB = √(x2 - x1)² + (y2 - y1)²

→ 10 = √(10 - 2)² + (x - (-3))²

→ 10 = √(8)² + (x + 3)²

→ 10 = √64 + x² + 9 + 6x

→ 10 = √73 + x² + 6x

• Squaring both side

→ 100 = 73 + x² + 6x

→ 100 - 73 = x² + 6x

→ 27 = x² + 6x

→ x² + 6x - 27 = 0

• Split middle of term

→ x² + 9x - 3x - 27 = 0

→ x(x + 9) - 3(x + 9) = 0

→ (x + 9)(x - 3) = 0

Either

→ x + 9 = 0

→ x = - 9

Or

→ x - 3 = 0

→ x = 3

Hence,

• The value of x is - 9 or 3

Answer 2

ɢɪᴠᴇɴ:-

• The distance between (2,-3) and (10,x) is 10 units

ᴛᴏ ғɪɴᴅ:-

• The value of "x".

ᴇxᴘʟᴀɴᴀᴛɪᴏɴ:-

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

x =$\sqrt{(a-c)^2+(b-d)^2}$

$\begin{lgathered}\begin{lgathered}\tt {\pink{Given}}\begin{cases} \sf{\green{A= (2, -3)}}\\ \sf{\blue{B = (10, x)}}\\ \sf{\orange{AB = 10units.}}\end{cases}\end{lgathered} \:\end{lgathered}$

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

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

10 =$\sqrt{(8)^2+(x+3)^2}$

10 =$\sqrt{64+(x)^2+9+6x}$

10 =$\sqrt{73+(x)^2+6x}$

Squaring both the sides,

100 = 73 + x² + 6x

100 - 73 = x² + 6x

27 = x² + 6x

x² + 6x - 27 = 0

[Splitting middle of term]

x² + 9x - 3x - 27 = 0

x(x + 9) - 3(x + 9) = 0

(x + 9)(x - 3) = 0

So,

x + 9 = 0

x = - 9

Or,

x - 3 = 0

x = 3

Therefore,

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

━━━━━━━━━━━━━━━━━━━━━━

✿$\huge\quad\underline\mathfrak\pink{Extra\:\:Bites:}$

✰ 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 :

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

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

internally in the ratio m₁ : m₂ is :

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

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

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

✰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:-

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