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Question: Find the value of \[x\] so that the points \[\left( {x, - 1} \right),\left( {2,1} \right)\] and \[\l...

Find the value of xx so that the points (x,1),(2,1)\left( {x, - 1} \right),\left( {2,1} \right) and (4,5)\left( {4,5} \right) are collinear.

Explanation

Solution

Here, we will use the slope of the two points formula to find the slope of any two line segments. We will equate the slope of both the line segments by using the condition of collinearity to find the value of the variable. If two or more points lie on a line close to each other or far away from each other, then the points are said to be collinear.

Formula Used:
Slope of the line segment mm joining two points say (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) is given by the formula: m=y2y1x2x1m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}

Complete step by step solution:
We are given that the points (x,1),(2,1)\left( {x, - 1} \right),\left( {2,1} \right) and (4,5)\left( {4,5} \right) are collinear.
Let A (x,1)\left( {x, - 1} \right), B (2,1)\left( {2,1} \right) and C (4,5)\left( {4,5} \right) be the given points.
We will first find the slope of the line segment.
Now, by using the slope of the two points formula m=y2y1x2x1m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}, we get slope of the line segment A (x,1)\left( {x, - 1} \right)and B(2,1)\left( {2,1} \right) as:
m1=1(1)2x{m_1} = \dfrac{{1 - \left( { - 1} \right)}}{{2 - x}}
Adding the terms in the numerator, we get
m1=22x\Rightarrow {m_1} = \dfrac{2}{{2 - x}} ……………………….(1)\left( 1 \right)
Now, by using the slope of the two points formula, we get slope of the line segment B (2,1)\left( {2,1} \right) and C (4,5)\left( {4,5} \right) as:
m2=5142{m_2} = \dfrac{{5 - 1}}{{4 - 2}}
Subtracting the terms, we get
m2=42\Rightarrow {m_2} = \dfrac{4}{2}
Dividing 4 by 2, we get
m2=2\Rightarrow {m_2} = 2 ………………………………(2)\left( 2 \right)
We know that if the given points are collinear then their slopes must be equal.
Now, by equating equation (1)\left( 1 \right) and equation (2)\left( 2 \right), we get
m1=m2{m_1} = {m_2}
22x=2\Rightarrow \dfrac{2}{{2 - x}} = 2
On cross multiplication, we get
2=2(2x)\Rightarrow 2 = 2\left( {2 - x} \right)
Multiplying the terms, we get
2=42x\Rightarrow 2 = 4 - 2x
Now, by rewriting the equation, we get
2x=42\Rightarrow 2x = 4 - 2
2x=2\Rightarrow 2x = 2
Dividing both sides by 2, we get
x=1\Rightarrow x = 1

Therefore, the value of xx is 11.

Note:
If three or more points are given then the given points are said to be collinear if the slope of any two pairs of points is the same. This can also be found by using the area of the triangle formula. If the area of the triangle formed by three points is zero, then they are said to be collinear.
Area of the triangle using three points is given by the formula A=12[x1(y2y3)+x2(y3y1)+x3(y1y2)]A = \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].
Let A (x,1)\left( {x, - 1} \right), B(2,1)\left( {2,1} \right) and C(4,5)\left( {4,5} \right) be the given points.
We are given that the points are collinear.
12[x1(y2y3)+x2(y3y1)+x3(y1y2)]=0\Rightarrow \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] = 0
By substituting the points, we get
12[x(15)+2(5(1))+4((1)1)]=0\Rightarrow \dfrac{1}{2}\left[ {x\left( {1 - 5} \right) + 2\left( {5 - \left( { - 1} \right)} \right) + 4\left( {\left( { - 1} \right) - 1} \right)} \right] = 0
Now, by simplifying the equation, we get
12[x(4)+2(6)+4(2)]=0\Rightarrow \dfrac{1}{2}\left[ {x\left( { - 4} \right) + 2\left( 6 \right) + 4\left( { - 2} \right)} \right] = 0
Multiplying the terms, we get
12[4x+128]=0\Rightarrow \dfrac{1}{2}\left[ { - 4x + 12 - 8} \right] = 0
12[4x+4]=0\Rightarrow \dfrac{1}{2}\left[ { - 4x + 4} \right] = 0
Multiplying both sides by 2, we get
4x+4=0\Rightarrow - 4x + 4 = 0
4x=4\Rightarrow - 4x = - 4
Dividing both sides by 4 - 4, we get
x=1\Rightarrow x = 1
Therefore, the value of xx is 11.