Answer
Verified
457.8k+ views
Hint: We will let the point on the x-axis be (x, 0) as the point is equidistant from (– 2, 5) and (2, – 3). So, we will calculate the distance of the point (x, 0) to each of the two points and then compare them to find the value of x. Once we get that, we have the point on the x-axis.
Complete step-by-step answer:
We are asked to find the point on the x-axis which is equidistant from the point (– 2, 5) and (2, – 3). We know that the point on the x-axis has its y – coordinate as 0. So, let (x, 0) be the point on the x-axis which is equidistant from (– 2, 5) and (2, – 3).
The distance between the two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is given by the distance formula \[D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] as we are given that (x, 0) is equidistant from (– 2, 5) and (2, – 3).
Now we will find their distance and then compare them to find our value of x.
Let \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( x,0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( -2,5 \right)\]
So, the distance between them will be given as,
\[{{D}_{1}}=\sqrt{{{\left( -2-x \right)}^{2}}+{{\left( 5-0 \right)}^{2}}}\]
We can write \[{{\left( -2-x \right)}^{2}}\] as
\[{{\left( -2-x \right)}^{2}}={{\left[ \left( - \right)\left( 2+x \right) \right]}^{2}}={{\left( 2+x \right)}^{2}}\]
So simplifying further, we get,
\[\Rightarrow {{D}_{1}}=\sqrt{{{\left( 2+x \right)}^{2}}+{{5}^{2}}}\]
Similarly, we can write, \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( x,0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( 2,-3 \right).\]
So, the distance between them will be given as,
\[{{D}_{2}}=\sqrt{{{\left( 2-x \right)}^{2}}+{{\left( -3-0 \right)}^{2}}}\]
So simplifying further, we get,
\[\Rightarrow {{D}_{2}}=\sqrt{{{\left( 2-x \right)}^{2}}+{{3}^{2}}}\]
As (x, 0) is equidistant from (– 2, 5) and (2, – 3), that means \[{{D}_{1}}={{D}_{2}}.\]
So, we get,
\[\sqrt{{{\left( 2+x \right)}^{2}}+{{5}^{2}}}=\sqrt{{{\left( 2-x \right)}^{2}}+{{3}^{2}}}\]
Squaring both the sides, we get,
\[{{\left( 2+x \right)}^{2}}+{{5}^{2}}={{\left( 2-x \right)}^{2}}+{{3}^{2}}\]
Opening the square brackets to simplify, we get,
\[\Rightarrow {{x}^{2}}+4+4x+25={{x}^{2}}+4-4x+9\]
Cancelling the like terms, we get,
\[\Rightarrow 8x=9-25\]
\[\Rightarrow 8x=-16\]
Dividing both the sides by 8, we get,
\[\Rightarrow \dfrac{8x}{8}=\dfrac{-16}{8}\]
\[\Rightarrow x=-2\]
So, we get, x = – 2 which means that the point on the x-axis that is equidistant from (– 2, 5) and (2, – 3) is (– 2, 0).
Hence, the required answer is (– 2, 0).
Note: We can cross-check that our solution is correct or not by the following steps. We will find the distance between (– 2, 0) and the other two points and see if they are equal or not.
(i) Distance between (– 2, 0) and (– 2, 5).
Let \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( -2,0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( -2,5 \right)\]
\[{{D}_{1}}=\sqrt{{{\left( -2-\left( -2 \right) \right)}^{2}}+{{\left( 5-0 \right)}^{2}}}\]
\[\Rightarrow {{D}_{1}}=\sqrt{0+{{5}^{2}}}\]
\[\Rightarrow {{D}_{1}}=5\]
We get the distance between (– 2, 0) and (– 2, 5) as 5 units.
(ii) Distance between (– 2, 0) and (2, – 3).
Let \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( -2,0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( 2,-3 \right)\]
\[{{D}_{2}}=\sqrt{{{\left( 2-\left( -2 \right) \right)}^{2}}+{{\left( -3-0 \right)}^{2}}}\]
\[\Rightarrow {{D}_{2}}=\sqrt{{{4}^{2}}+{{3}^{2}}}\]
\[\Rightarrow {{D}_{2}}=\sqrt{16+9}\]
\[\Rightarrow {{D}_{2}}=\sqrt{25}\]
\[\Rightarrow {{D}_{2}}=5\]
So again, we get the distance between (– 2, 0) and (– 2, 5) as 5 units.
So, our answer is correct.
Complete step-by-step answer:
We are asked to find the point on the x-axis which is equidistant from the point (– 2, 5) and (2, – 3). We know that the point on the x-axis has its y – coordinate as 0. So, let (x, 0) be the point on the x-axis which is equidistant from (– 2, 5) and (2, – 3).
The distance between the two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is given by the distance formula \[D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] as we are given that (x, 0) is equidistant from (– 2, 5) and (2, – 3).
Now we will find their distance and then compare them to find our value of x.
Let \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( x,0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( -2,5 \right)\]
So, the distance between them will be given as,
\[{{D}_{1}}=\sqrt{{{\left( -2-x \right)}^{2}}+{{\left( 5-0 \right)}^{2}}}\]
We can write \[{{\left( -2-x \right)}^{2}}\] as
\[{{\left( -2-x \right)}^{2}}={{\left[ \left( - \right)\left( 2+x \right) \right]}^{2}}={{\left( 2+x \right)}^{2}}\]
So simplifying further, we get,
\[\Rightarrow {{D}_{1}}=\sqrt{{{\left( 2+x \right)}^{2}}+{{5}^{2}}}\]
Similarly, we can write, \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( x,0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( 2,-3 \right).\]
So, the distance between them will be given as,
\[{{D}_{2}}=\sqrt{{{\left( 2-x \right)}^{2}}+{{\left( -3-0 \right)}^{2}}}\]
So simplifying further, we get,
\[\Rightarrow {{D}_{2}}=\sqrt{{{\left( 2-x \right)}^{2}}+{{3}^{2}}}\]
As (x, 0) is equidistant from (– 2, 5) and (2, – 3), that means \[{{D}_{1}}={{D}_{2}}.\]
So, we get,
\[\sqrt{{{\left( 2+x \right)}^{2}}+{{5}^{2}}}=\sqrt{{{\left( 2-x \right)}^{2}}+{{3}^{2}}}\]
Squaring both the sides, we get,
\[{{\left( 2+x \right)}^{2}}+{{5}^{2}}={{\left( 2-x \right)}^{2}}+{{3}^{2}}\]
Opening the square brackets to simplify, we get,
\[\Rightarrow {{x}^{2}}+4+4x+25={{x}^{2}}+4-4x+9\]
Cancelling the like terms, we get,
\[\Rightarrow 8x=9-25\]
\[\Rightarrow 8x=-16\]
Dividing both the sides by 8, we get,
\[\Rightarrow \dfrac{8x}{8}=\dfrac{-16}{8}\]
\[\Rightarrow x=-2\]
So, we get, x = – 2 which means that the point on the x-axis that is equidistant from (– 2, 5) and (2, – 3) is (– 2, 0).
Hence, the required answer is (– 2, 0).
Note: We can cross-check that our solution is correct or not by the following steps. We will find the distance between (– 2, 0) and the other two points and see if they are equal or not.
(i) Distance between (– 2, 0) and (– 2, 5).
Let \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( -2,0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( -2,5 \right)\]
\[{{D}_{1}}=\sqrt{{{\left( -2-\left( -2 \right) \right)}^{2}}+{{\left( 5-0 \right)}^{2}}}\]
\[\Rightarrow {{D}_{1}}=\sqrt{0+{{5}^{2}}}\]
\[\Rightarrow {{D}_{1}}=5\]
We get the distance between (– 2, 0) and (– 2, 5) as 5 units.
(ii) Distance between (– 2, 0) and (2, – 3).
Let \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( -2,0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( 2,-3 \right)\]
\[{{D}_{2}}=\sqrt{{{\left( 2-\left( -2 \right) \right)}^{2}}+{{\left( -3-0 \right)}^{2}}}\]
\[\Rightarrow {{D}_{2}}=\sqrt{{{4}^{2}}+{{3}^{2}}}\]
\[\Rightarrow {{D}_{2}}=\sqrt{16+9}\]
\[\Rightarrow {{D}_{2}}=\sqrt{25}\]
\[\Rightarrow {{D}_{2}}=5\]
So again, we get the distance between (– 2, 0) and (– 2, 5) as 5 units.
So, our answer is correct.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
Write the difference between order and molecularity class 11 maths CBSE
A rainbow has circular shape because A The earth is class 11 physics CBSE
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What are noble gases Why are they also called inert class 11 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between calcination and roasting class 11 chemistry CBSE