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Question: Find the relation between \[x\] and \[y\] if the points \[A\left( {x,y} \right),B\left( { - 5,7} \ri...

Find the relation between xx and yy if the points A(x,y),B(5,7)A\left( {x,y} \right),B\left( { - 5,7} \right) and C(4,5)C\left( { - 4,5} \right) are collinear.

Explanation

Solution

We need to find the relation between xx and yy as all the given points are collinear. We know that the points A,BA,B and CC are collinear then area of ΔABC=0\Delta ABC = 0. So, we will calculate area of triangle using 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] and put it equal to 0 to find the relation between xx and yy.

Complete step by step solution: We will first consider the given points A(x,y),B(5,7)A\left( {x,y} \right),B\left( { - 5,7} \right) and C(4,5)C\left( { - 4,5} \right).
We need to find the relation between xx and yy if the given points are collinear.
Now, we know that the points A(x,y),B(5,7)A\left( {x,y} \right),B\left( { - 5,7} \right) and C(4,5)C\left( { - 4,5} \right) are collinear then area of ΔABC=0\Delta ABC = 0.
Thus, we will use 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] to find the area of triangle where (x1,y1);(x2,y2)\left( {{x_1},{y_1}} \right);\left( {{x_2},{y_2}} \right) and (x3,y3)\left( {{x_3},{y_3}} \right) are the coordinated of A,BA,B and CC.
Hence, we will substitute the value of coordinates given in the formula and find the value of area of triangle,
Thus, we get,

A=12[x(75)5(5y)4(y7)] A=12[2x25+5y4y+28] A=12[2x+y+3]  \Rightarrow A = \dfrac{1}{2}\left[ {x\left( {7 - 5} \right) - 5\left( {5 - y} \right) - 4\left( {y - 7} \right)} \right] \\\ \Rightarrow A = \dfrac{1}{2}\left[ {2x - 25 + 5y - 4y + 28} \right] \\\ \Rightarrow A = \dfrac{1}{2}\left[ {2x + y + 3} \right] \\\

Now, as we know that the points A,BA,B and CC are collinear then area of ΔABC=0\Delta ABC = 0.
Thus, we get,

12[2x+y+3]=0 2x+y+3=0 2x+y=3  \Rightarrow \dfrac{1}{2}\left[ {2x + y + 3} \right] = 0 \\\ \Rightarrow 2x + y + 3 = 0 \\\ \Rightarrow 2x + y = - 3 \\\

Hence, we get the relation between xx and yy is 2x+y=32x + y = - 3.

Note: We must remember that the coordinates are collinear when the area of the triangle is equal to zero and using this fact only, we have found the relation between xx and yy. We have to remember 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] to determine the area of triangle using the coordinates of the triangle. We can easily determine the relation as one of the coordinates are given as (x,y)\left( {x,y} \right).