Question

Find the value of 'x' if the distance between the points (4,-3) and (-2,x) is 10cm

Answer 1

$\bold{\boxed{\begin{minipage}{9 cm}Basic Formulas \\ \\Distance\; Formula =\sqrt{(x_{2} - x_{1} )^{2} +(y_{2}  - y_{1} )^{2} } \\ \\Mid\; Point\; Formula = [\frac{x_{1}+x_{2}  }{2},\frac{y_{1}+y_{2}  }{2}]\\ \\Section\; Formula = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+mx_{2}  },\frac{m_{1}y_{2} +m_{2}y_{1}  }{m_{1}+m_{2}  } \\  \end{minipage}}}$

$\mathfrak{Question:-}$

Find the value of 'x' if the distance between the points (4,-3) and (-2,x) is 10cm.

$\mathfrak{Answer:-}$

$\small{\textsf{\underline{\underline{Given:-}}}}$

Let two points are A and B.

A = (4, -3)

B = (-2 , x)

Distance between AB = 10cm.

$\large{\textsf{\underline{\underline{By\; Distance\; Formula:-}}}}$

$\bold{= \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }}$

$AB=\sqrt{(-2-4)^{2}+(x-(-3))^{2}$

$10 = \sqrt{(-6)^{2}+(x+3)^{2}  }$

$10 = \sqrt{36+x^{2}+9+6x }$

Squaring Both the sides, We get

$\bold{100 = 36+9+x^{2}+6x}$

$\bold{100 = 45+x^{2}+6x}$

$\bold{= x^{2}+6x = 55}$

By Using Splitting Middle term method

$\bold{= x^{2} +6x -55 = 0}$

$\bold{= x^{2} +11x-5x-55=0}$

$\bold{= x(x+11)-5(x+11)}$

$\bold{= (x-5) (x+11)}$

$\bold{= x= 5 \;or\; x = -11}$

Answer 2

$\huge\bf\pink{\mid{\overline{\underline{Your\: Answer}}}\mid}$

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${\bold{\red{\huge{x\:=\:5,\:-11}}}}$

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step-by-step explanation :

We know that,

If two points,

A(x1,y1) and B(x2,y2) is given,

then distance between these points,

is given by the Distance Formula,

D = √{${(x2-x1)}^{2}$+${(y2-y1)}^{2}$}

Now,

Let the given pts. be

A (4,-3)

B (-2,x)

Also,

it is given that,

Distance between these points,D = 10 cm

So,

putting these values in Distance Formula,

we get,

√{${(-2-4)}^{2}$+${[x-(-3)]}^{2}$} = 10

Squaring on both sides,

we get,

=> ${(-6)}^{2}$+${(x+3)}^{2}$ = ${10}^{2}$

=> 36 + ${(x+3)}^{2}$= 100

Subtracting 36 from both sides,

we get,

=> ${(x+3)}^{2}$ = 64

=> x + 3 = ±8

=> x = 8 - 3 or x = -8 - 3

=> x = 5 or x = -11


Hence,

x = 5 and x = -11