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Question: A, B and C are three collinear points. The Coordinates of A and B are \(\left( {3,4} \right)\) and \...

A, B and C are three collinear points. The Coordinates of A and B are (3,4)\left( {3,4} \right) and (7,7)\left( {7,7} \right) respectively and AC=10AC = 10 Units. Find the sum of coordinates of C.

Explanation

Solution

We use the midpoint formula to solve this problem and first find out the coordinates of C and their sum.

Formula Used:
Coordinates of the middle point G of two points M(x,y)M\left( {x,y} \right) and N(a,b)N\left( {a,b} \right) which lie of the same line is given by G(x+a2,y+b2)G\left( {\dfrac{{x + a}}{2},\dfrac{{y + b}}{2}} \right).

Complete step-by-step answer:
A,B and C collinear points, as they lie on the same line.

Let the coordinate of point C be (x,y)\left( {x,y} \right)
Distance formula between two points (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) , d=(x2x1)2+(y2y1)2d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}
Using the distance formula between two points A(3,4)A\left( {3,4} \right) and B(7,7)B\left( {7,7} \right)

dAB=(73)2+(74)2 dAB=16+9 dAB=5  {d_{AB}} = \sqrt {{{\left( {7 - 3} \right)}^2} + {{\left( {7 - 4} \right)}^2}} \\\ {d_{AB}} = \sqrt {16 + 9} \\\ {d_{AB}} = 5 \\\

The distance between the points A and B is 5 units. The distance between B and C is given by
dBC=dACdAB dBC=105 dBC=5  {d_{BC}} = {d_{AC}} - {d_{AB}} \\\ {d_{BC}} = 10 - 5 \\\ {d_{BC}} = 5 \\\
It means that B is the mid-point of A and C.
The coordinates of the mid-point G, lying between the two points M(x,y)M\left( {x,y} \right) and N(a,b)N\left( {a,b} \right) , is given by G(x+a2,y+b2)G\left( {\dfrac{{x + a}}{2},\dfrac{{y + b}}{2}} \right).
Using it, find the coordinates of point B.
Coordinates of A(3,4)A\left( {3,4} \right) and C(x,y)C\left( {x,y} \right)
B(3+x2,4+y2)B\left( {\dfrac{{3 + x}}{2},\dfrac{{4 + y}}{2}} \right)
But the coordinates of B is (7,7)\left( {7,7} \right) . Equate the X-Coordinate and Y-Coordinate of B(3+x2,4+y2)B\left( {\dfrac{{3 + x}}{2},\dfrac{{4 + y}}{2}} \right) and B(7,7)B\left( {7,7} \right) to calculate the coordinates of C(x,y)C\left( {x,y} \right)
3+x2=7 x=11  \dfrac{{3 + x}}{2} = 7 \\\ x = 11 \\\
Also,
4+y2=7 y=10  \dfrac{{4 + y}}{2} = 7 \\\ y = 10 \\\
The coordinates of point C is (11,10)\left( {11,10} \right) .
The sum of coordinates of point C,
S=11+10 S=21  S = 11 + 10 \\\ S = 21 \\\
Thus, the sum of the coordinates of point C is S=21S = 21

Note: Collinear points are those points which lie on the same line. For A, B, and C to be collinear, the area of the triangle should be equal to 0.
Important concept and formulas to be remembered are
The distance between two points (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) , d=(x2x1)2+(y2y1)2d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}
The midpoint of two points, all lying on the same line is calculated as , G(x+a2,y+b2)G\left( {\dfrac{{x + a}}{2},\dfrac{{y + b}}{2}} \right) where two points are M(x,y)M\left( {x,y} \right) and N(a,b)N\left( {a,b} \right).