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Question: If \(P, Q, R\) are collinear points such that \(P\left( {3,4} \right)\), \(Q\left( {7,7} \right)\) a...

If P,Q,RP, Q, R are collinear points such that P(3,4)P\left( {3,4} \right), Q(7,7)Q\left( {7,7} \right) and PR=10PR = 10, find RR.

Explanation

Solution

Since we have two coordinate points we consider the coordinate R as (a,b)\left( {a,b} \right). First, we find the distance between PP and RR. We also know that if the three points are collinear then the area of the triangle formed by the three points must be equal to zero. So, we must remember the formula for the area of the triangle given the coordinates of the three points of the triangle.

Complete step by step answer:
Let the x-coordinate of point R be “a” and the y-coordinate of point R be “b”.
So, the coordinates of point R are: (a,b)\left( {a,b} \right)
The distance formula between any two points, say (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) is written as:
Distance between two points =(x2−x1)2+(y2−y1)2= \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}
The distance between points P and R is calculated as:
PR=(a−3)2+(b−4)2= \sqrt {{{\left( {a - 3} \right)}^2} + {{\left( {b - 4} \right)}^2}}
The length of side PR is equal to 1010 units.
By equalizing distance PR with its value, we get,
(a−3)2+(b−4)2=10\sqrt {{{\left( {a - 3} \right)}^2} + {{\left( {b - 4} \right)}^2}} = 10
Squaring on both sides we have
⇒(a−3)2+(b−4)2=(10)2\Rightarrow {\left( {a - 3} \right)^2} + {\left( {b - 4} \right)^2} = {\left( {10} \right)^2}
⇒(a−3)2+(b−4)2=100\Rightarrow {\left( {a - 3} \right)^2} + {\left( {b - 4} \right)^2} = 100
Now, using the algebraic identity (a−b)2=a2−2ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}, we get,
⇒a2−2(a)(3)+32+b2−2(b)(4)+42=100\Rightarrow {a^2} - 2\left( a \right)\left( 3 \right) + {3^2} + {b^2} - 2\left( b \right)\left( 4 \right) + {4^2} = 100
⇒a2−6a+9+b2−8b+16=100\Rightarrow {a^2} - 6a + 9 + {b^2} - 8b + 16 = 100
⇒a2−6a+b2−8b=75−−−−(1)\Rightarrow {a^2} - 6a + {b^2} - 8b = 75 - - - - \left( 1 \right)

As P, Q and R are collinear. So, the area of triangle PQR should be equal to zero as collinear points cannot form any triangle. That is if three points (x1,y1)\left( {{x_1},{y_1}} \right), (x2,y2)\left( {{x_2},{y_2}} \right) and (x3,y3)\left( {{x_3},{y_3}} \right).
Now, we know that area of triangle with coordinates of vertices as (x1,y1)\left( {{x_1},{y_1}} \right), (x2,y2)\left( {{x_2},{y_2}} \right) and (x3,y3)\left( {{x_3},{y_3}} \right) is 12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\dfrac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|.
So, area of triangle PQR =12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣ = \dfrac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|
Now, substituting the values of coordinates of points P, Q and R, we get,
⇒12∣3(7−b)+7(b−4)+a(4−7)∣=0\Rightarrow \dfrac{1}{2}\left| {3\left( {7 - b} \right) + 7\left( {b - 4} \right) + a\left( {4 - 7} \right)} \right| = 0
Opening the brackets and simplifying the expression, we get,
⇒12∣21−3b+7b−28+4a−7a∣=0\Rightarrow \dfrac{1}{2}\left| {21 - 3b + 7b - 28 + 4a - 7a} \right| = 0
⇒12∣−7+4b−3a∣=0\Rightarrow \dfrac{1}{2}\left| { - 7 + 4b - 3a} \right| = 0
So, we get the value of (−7+4b−3a)\left( { - 7 + 4b - 3a} \right) as zero.

Now, (−7+4b−3a)=0−−−−(2)\left( { - 7 + 4b - 3a} \right) = 0 - - - - \left( 2 \right).
So, we have to solve the equations (1)\left( 1 \right) and (2)\left( 2 \right) to get the values of a and b.
We get the value of a as (4b−7)3\dfrac{{\left( {4b - 7} \right)}}{3} from equation (2)\left( 2 \right). So, substituting this in equation (1)\left( 1 \right), we get,
⇒(4b−73)2−6(4b−73)+b2−8b=75\Rightarrow {\left( {\dfrac{{4b - 7}}{3}} \right)^2} - 6\left( {\dfrac{{4b - 7}}{3}} \right) + {b^2} - 8b = 75
⇒(4b−7)29−2(4b−7)+b2−8b=75\Rightarrow \dfrac{{{{\left( {4b - 7} \right)}^2}}}{9} - 2\left( {4b - 7} \right) + {b^2} - 8b = 75
Now, computing the square using the algebraic identity (a−b)2=a2−2ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}, we get,
⇒16b2−56b+499−8b+14+b2−8b=75\Rightarrow \dfrac{{16{b^2} - 56b + 49}}{9} - 8b + 14 + {b^2} - 8b = 75
Now, taking LCM of the denominator, we get,
⇒16b2−56b+49−72b+126+9b2−72b=675\Rightarrow 16{b^2} - 56b + 49 - 72b + 126 + 9{b^2} - 72b = 675
⇒25b2−200b+175=675\Rightarrow 25{b^2} - 200b + 175 = 675
⇒25b2−200b−500=0\Rightarrow 25{b^2} - 200b - 500 = 0
Now, dividing both sides of the equation by 2525.
⇒b2−8b−20=0\Rightarrow {b^2} - 8b - 20 = 0

Now, we split the middle term of the quadratic equation to find the value of b.
⇒b2−10b+2b−20=0\Rightarrow {b^2} - 10b + 2b - 20 = 0
⇒b(b−10)+2(b−10)=0\Rightarrow b\left( {b - 10} \right) + 2\left( {b - 10} \right) = 0
⇒(b+2)(b−10)=0\Rightarrow \left( {b + 2} \right)\left( {b - 10} \right) = 0
Now, either (b+2)=0\left( {b + 2} \right) = 0 or (b−10)=0\left( {b - 10} \right) = 0
Either b=−2b = - 2 or b=10b = 10
Now, putting the values of b in equation (2)\left( 2 \right) to find the values of a, we get,
⇒a=(4b−7)3\Rightarrow a = \dfrac{{\left( {4b - 7} \right)}}{3}
For b=10b = 10,
⇒a=(4×10−7)3\Rightarrow a = \dfrac{{\left( {4 \times 10 - 7} \right)}}{3}
⇒a=11\Rightarrow a = 11
For b=−2b = - 2,
⇒a=(4×(−2)−7)3\Rightarrow a = \dfrac{{\left( {4 \times \left( { - 2} \right) - 7} \right)}}{3}
∴a=−153=−5\therefore a = \dfrac{{ - 15}}{3} = - 5

So, the coordinates of Point R are: (−5,−2)\left( { - 5, - 2} \right) and (11,10)\left( {11,10} \right).

Note: We must take care of calculations while doing such questions so as to be sure of the final answer. We must remember the distance formula to find the length of PRPR. We should also know the formula for the area of the triangle when we are given the coordinates of the vertices. A modulus function is used in the formula of the area of the triangle as the area must be positive. We must be patient while solving such problems as it consists of lengthy calculations.