Distance between two points (X,7) and (1,15) is 10 units.
To Find :
value of x.
We can find the value of x using Distance Formula. As it is given distance between two points is 10 Units and two points are (x ,7) and (1,15). So we will substitute the values in distance formula to get value of x.
[tex] \red\bigstar\:\boxed{\sf AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}} [/tex]
Question:
[tex]\\[/tex]
The distance between two points (x, 7) and (1,15) is 10 units.Find the value of x .
Solution:
To find: Find the value of x.
Step-by-step explanation:
❍Let us consider that, [tex]\rm{x_1 = x,x_2 =1}[/tex] and [tex]\rm{y_1 = 7,y_2 =15}[/tex]
As we know that,
[tex] \sf{Distance \: between \: two \: points \: AB = \sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}}[/tex]
Applying the values on given formula,
[tex] \longrightarrow\sf{10 = \sqrt{ {(1 – x)}^{2} + {(15 -7 )}^{2} } } \\\longrightarrow\sf{10 = \sqrt{{(1 – x)}^{2} + {(8)}^{2}}} \\ \longrightarrow\sf{ 10 = \sqrt{{(1 – x)}^{2} + 64}} \\ \longrightarrow \sf{ {10}^{2} = {(1 – x)}^{2} + 64}\\\longrightarrow\sf{100 = {(1 – x)}^{2} + 64 } \\ \longrightarrow \sf{100 – 64 = {(1 – x)}^{2}} \\ \longrightarrow \sf{36 = {(1 – x)}^{2}} \\ \longrightarrow \sf{ \sqrt{36} = (1 – x) } \\ \longrightarrow \sf{ \pm \: 6 = (1 – x)} \\ \longrightarrow \sf{x = \pm \: 6 – 1} \\ \longrightarrow \sf{x = – 5(or)7}[/tex]
________________________
Answer :
Solution :
Given :
To Find :
We can find the value of x using Distance Formula. As it is given distance between two points is 10 Units and two points are (x ,7) and (1,15). So we will substitute the values in distance formula to get value of x.
[tex] \red\bigstar\:\boxed{\sf AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}} [/tex]
Here,
[tex]\LARGE \color{blue}\mathfrak{Substituting\:the\:values}[/tex]
[tex]\implies\sf 10 = \sqrt{(1-x)^2+(15-7)^2} [/tex]
[tex]\implies\sf 10 = \sqrt{(1-x)^2+(8)^2}\\ [/tex]
• Squaring both the sides
[tex]\implies\sf 10^2 = (1-x)^2+(8)^2\\ [/tex]
[tex]\implies\sf 100 = (1-x)^2+64\\ [/tex]
[tex]\implies\sf 36 = (1-x)^2\\ [/tex]
• Square root both the sides
[tex]\implies\sf \sqrt{36} = \sqrt{(1-x)^2}\\ [/tex]
[tex]\implies\sf ±6 = 1-x\\ [/tex]
[tex]\implies \sf 1-x = ±6 \\ [/tex]
[tex]\implies \sf (1-x) = 6 \:\: or \:\: (1-x) = -6 \\ [/tex]
[tex]\implies \sf 1-x = 6 \:\: or \:\: 1-x = -6 \\ [/tex]
[tex] \implies \sf -x = 6-1 \:\: or\; \: -x = -6-1 \\ [/tex]
[tex]\implies \sf -x = 5 \:\: or \:\: -x = -7 \\ \\ \implies \sf \underline{\boxed{\pink{\sf x = -5 \: \: or\: \: 7 }}}[/tex]
Hence,
value of x is 7 or -5.
[tex]{\fcolorbox{red}{blue}{\orange{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: SugarCrash\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}} [/tex]