Question

A line of length 10 and one end is at the point (2, – 3); if the abscissa of the other end is 10, prove that its ordinate must be 3 or – 9.

Answer 1

Hint: First of all, assume a line segment AB of 10 units in which we take coordinates of A as (2, – 3) and coordinates of B as (10, y). Now apply distance formula that is \[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] between points A and B and equate it to 10 to find the required value of y.

Complete step by step answer:
Here, we are given that a line segment is of the length 10 and one end is at the point (2, – 3) if the abscissa of the other end is 10, we have to prove that its ordinate must be 3 or – 9. First of all, let us consider a line segment AB in which coordinates of point A is (2, – 3). Also, let us take the coordinate of point B as (10, y).

Now, we know that the distance between two points is given by:
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}.....\left( i \right)\]
Since we already know the distance between coordinates A and B is 10 units. We also, know the coordinates of point A and abscissa of point B, by substituting the value, we know in equation (i), we get,
\[10=\sqrt{{{\left( 10-2 \right)}^{2}}+{{\left( y-\left( -3 \right) \right)}^{2}}}\]
\[10=\sqrt{{{\left( 8 \right)}^{2}}+{{\left( y+3 \right)}^{2}}}\]
\[10=\sqrt{64+{{\left( y+3 \right)}^{2}}}\]
By squaring both the sides of the above equation, we get,
\[{{\left( 10 \right)}^{2}}=64+{{\left( y+3 \right)}^{2}}\]
\[100=64+{{\left( y+3 \right)}^{2}}\]
By subtracting 64 from both sides of the above equation, we get,
\[{{\left( y+3 \right)}^{2}}=100-64\]
\[\Rightarrow {{\left( y+3 \right)}^{2}}=36\]
We know that, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. By using this formula in LHS of the above equation, we get,
\[{{y}^{2}}+{{\left( 3 \right)}^{2}}+2.y.3=36\]
\[\Rightarrow {{y}^{2}}+9+6y-36=0\]
\[\Rightarrow {{y}^{2}}+6y-27=0\]
We can also write the above equation as,
\[{{y}^{2}}+9y-3y-27=0\]
\[\Rightarrow y\left( y+9 \right)-3\left( y+9 \right)=0\]
By taking out (y + 9) common from the above equation, we get,
\[\left( y+9 \right)\left( y-3 \right)=0\]
So, we get, y + 9 = 0 or y – 3 = 0
Hence, y = – 9 or y = 3
So, we have proved that the values of the ordinates of the point B must be 3 or – 9.

Note: Students must note that the length of the line means the distance between its points of extremities and will be calculated by using the distance formula. Also, apart from solving the quadratic equation at the end of the solution to find the values of y, we can take root to find the values of y in this way.
\[{{\left( y+3 \right)}^{2}}=36\]
\[\sqrt{{{\left( y+3 \right)}^{2}}}=\sqrt{36}\]
\[\Rightarrow y+3=\pm 6\]
\[y=\pm 6-3\]
So, we get y = 6 – 3 or y = – 6 – 3
Hence, we get, y = 3 or y = – 9.