Question

Find the value of x such that \[PQ=QR\] where the coordinates of \[P,Q\] and \[R\] are \[\left( 6,-1 \right)\]\[\left( 1,3 \right)\] and \[\left( x,8 \right)\] respectively.

Answer 1

Hint: To solve this type of problem first we have to know the distance formula between two points. Then applying the distance formula and equating both as given in the question and solving gives the value of x.

Complete step-by-step answer:
Given \[PQ=QR\]
The points given are \[P=\left( 6,-1 \right)\],\[Q=\left( 1,3 \right)\],\[R=\left( x,8 \right)\].
If the two points considered are \[A\left( {{x}_{1}},{{y}_{1}} \right)\]and \[B\left( {{x}_{2}},{{y}_{2}} \right)\]then the distance between two points is given by \[AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
In the question it was given that \[PQ=QR\], Now writing the distance between \[PQ\] and \[QR\].
\[PQ=\sqrt{{{\left( 1-6 \right)}^{2}}+{{\left( 3-(-1) \right)}^{2}}}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
\[QR=\sqrt{{{\left( x-1 \right)}^{2}}+{{\left( 8-3 \right)}^{2}}}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (b)
Now equating both (a) and (b) we get,
\[\sqrt{{{\left( 1-6 \right)}^{2}}+{{\left( 3-(-1) \right)}^{2}}}=\sqrt{{{\left( x-1 \right)}^{2}}+{{\left( 8-3 \right)}^{2}}}\]
Squaring on both sides we get,
\[{{\left( 1-6 \right)}^{2}}+{{\left( 3-(-1) \right)}^{2}}={{\left( x-1 \right)}^{2}}+{{\left( 8-3 \right)}^{2}}\]
\[25+16={{x}^{2}}-2x+1+25\]
\[{{x}^{2}}-2x=41-26\]
\[{{x}^{2}}-2x=15\]
\[{{x}^{2}}-2x-15=0\]
We have got quadratic equation we have to find the roots,
\[{{x}^{2}}-2x-15=0\]
\[{{x}^{2}}-5x+3x-15=0\]
\[x(x-5)+3(x-5)=0\]
\[\left( x-5 \right)\left( x+3 \right)=0\]
\[x=5,-3\]
The values of x are \[x=5,-3\].

Note: This is a direct problem, with knowing the formula of distance between two points we can solve this. The distance between any two points cannot be negative. If we get a negative value also we have to apply modulus. Take care while doing calculations.