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Question: How do you find the value of “a” given point \(\left( 4,-1 \right),\left( a,-5 \right)\) with a dist...

How do you find the value of “a” given point (4,−1),(a,−5)\left( 4,-1 \right),\left( a,-5 \right) with a distance of 1010 ?

Explanation

Solution

Since, the distance between two points are given. So, with use of the formula of distance between two points that is d=(x2−x1)2+(y2−y1)2d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} , we can get the value of the point aa by applying all the values in the formula.

Complete step-by-step solution:
Since, we have given two points as (4,−1)\left( 4,-1 \right) and (a,−5)\left( a,-5 \right) respectively and we also have the distance between these two points that is equal to 1010. Now, we will apply the formula and put the value as:
⇒d=(x2−x1)2+(y2−y1)2\Rightarrow d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}
Here, we will take (4,−1)\left( 4,-1 \right) and (a,−5)\left( a,-5 \right) as (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) respectively. Now, we will put the value as:
⇒10=(a−4)2+(−5−1)2\Rightarrow 10=\sqrt{{{\left( a-4 \right)}^{2}}+{{\left( -5-1 \right)}^{2}}}
Now, we will square both sides. The above step will be below as:
⇒(10)2=((a−4)2+(−5−1)2)2\Rightarrow {{\left( 10 \right)}^{2}}={{\left( \sqrt{{{\left( a-4 \right)}^{2}}+{{\left( -5-1 \right)}^{2}}} \right)}^{2}}
After squaring, the under root will be removed as:
⇒100=(a−4)2+(−5−1)2\Rightarrow 100={{\left( a-4 \right)}^{2}}+{{\left( -5-1 \right)}^{2}}
Now, the above step will be below as:
⇒100=(a−4)2+(−6)2\Rightarrow 100={{\left( a-4 \right)}^{2}}+{{\left( -6 \right)}^{2}}
Here, we will expand the above expression as:
⇒100=a2+42−2×a×4+(−6)2\Rightarrow 100={{a}^{2}}+{{4}^{2}}-2\times a\times 4+{{\left( -6 \right)}^{2}}
Now, we will solve by completing the square and multiplication in the above step as:
⇒100=a2+16−8a+36\Rightarrow 100={{a}^{2}}+16-8a+36
Here, we will change the place of 100100 and the above equation will be as:
⇒a2+16−8a+36−100=0\Rightarrow {{a}^{2}}+16-8a+36-100=0
Now, we will solve the numbers as:
⇒a2+16−8a−64=0\Rightarrow {{a}^{2}}+16-8a-64=0
After using subtraction for numbers, we will have the above step as:
⇒a2−8a−48=0\Rightarrow {{a}^{2}}-8a-48=0
Now, we will use the factorization to get the value of point aa . So, we can write −8a-8a as (12a+4a)\left( 12a+4a \right)in the above step as:
⇒a2−12a+4a−48=0\Rightarrow {{a}^{2}}-12a+4a-48=0
Here, we can take aa as a common factor for first two terms and 44 for last two terms as:
⇒a(a−12)+4(a−12)=0\Rightarrow a\left( a-12 \right)+4\left( a-12 \right)=0
Now, we can write the above step below as:
⇒(a+4)(a−12)=0\Rightarrow \left( a+4 \right)\left( a-12 \right)=0
Here, we will take first factor:
⇒a+4=0\Rightarrow a+4=0
We can get the value of aa as:
⇒a=−4\Rightarrow a=-4
Now, we will use second factor as:
⇒a−12=0\Rightarrow a-12=0
Thus, the value of aa will be as:
⇒a=12\Rightarrow a=12
Hence, we have the value of aa as −4-4 and 1212.

Note: Here, we will check if the solution is correct or not by putting any value of aain the following way:
Since, we have distance formula as:
⇒10=(a−4)2+(−5−1)2\Rightarrow 10=\sqrt{{{\left( a-4 \right)}^{2}}+{{\left( -5-1 \right)}^{2}}}
Now, we will put a=−4a=-4 as:
⇒10=(−4−4)2+(−5−1)2\Rightarrow 10=\sqrt{{{\left( -4-4 \right)}^{2}}+{{\left( -5-1 \right)}^{2}}}
Here, we will solve the bracketed terms as:
⇒10=(−8)2+(−6)2\Rightarrow 10=\sqrt{{{\left( -8 \right)}^{2}}+{{\left( -6 \right)}^{2}}}
Now, we will complete the square as:
⇒10=64+36\Rightarrow 10=\sqrt{64+36}
After getting the sum of 6464 and 3636 , we will have:
⇒10=100\Rightarrow 10=\sqrt{100}
As we know that the 1010 is square root of 100100 . So, the above equation will be as:
⇒10=10\Rightarrow 10=10
Since, L.H.S.=R.H.S.
Hence, the solution is correct.