Question

How do you find the value of “a” given point $\left( 4,-1 \right),\left( a,-5 \right)$ with a distance of $10$ ?

Answer 1

Hint: Since, the distance between two points are given. So, with use of the formula of distance between two points that is \[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] , we can get the value of the point $a$ by applying all the values in the formula.

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

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