Question

a) Verify whether the following points are collinear or not: (1, 4), (3, -2) and (-3, 16)
b) The area of the quadrilateral formed by (1, 1), (t, 3), (4, 6) and (-2, 5) is 13 sq. units then find t.

Answer 1

Hint: a) Assume A, B, C are three points. If AB+BC=AC or AB+AC=BC or BC+AC=AB then the points are said to be collinear. Find the distances between the points given in the question. If they satisfy any of these conditions then the points are said to be collinear.
b) Find the area of the quadrilateral using the formula when four vertices of quadrilateral are given and then equate the area with 13 to get the value of t.

Formulas used:
Distance between two points A (x1, y1) and B (x2, y2) is $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Area of a quadrilateral when four vertices are given is \[\dfrac{1}{2}\left\{ {\begin{array}{*{20}{c}}
  {{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_1}} \\
  {{y_1} \nearrow }&{ \searrow {y_2} \nearrow }&{ \searrow {y_3} \nearrow }&{ \searrow {y_4} \nearrow }&{ \searrow {y_1}}
\end{array}} \right\}\]
$ = \dfrac{1}{2}\left\{ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_4} + {x_4}{y_1}} \right) - \left( {{x_2}{y_1} + {x_3}{y_2} + {x_4}{y_3} + {x_1}{y_4}} \right)} \right\}$


Complete step-by-step Solution:
a) Given points (1, 4), (3, -2) and (-3, 16). We have to find whether these points are collinear or not.
Let A(x1, y1) = (1, 4), B(x2, y2) = (3, -2), C(x3, y3) = (-3, 16)
Distance between A and B= $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
\[\left( {x1,y1} \right) = \left( {1,4} \right),\left( {x2,y2} \right) = \left( {3, - 2} \right)\]
AB= $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
$
   = \sqrt {{{\left( {3 - 1} \right)}^2} + {{\left( { - 2 - 4} \right)}^2}} \\
   = \sqrt {{2^2} + {{\left( { - 6} \right)}^2}} \\
   = \sqrt {4 + 36} \\
   = \sqrt {40} \\
   = \sqrt {4 \times 10} \\
   = 2\sqrt {10} units \\
 $
Distance between B and C= $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
\[\left( {x1,y1} \right) = \left( {3, - 2} \right),\left( {x2,y2} \right) = \left( { - 3,16} \right)\]
BC= $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
$
   = \sqrt {{{\left( { - 3 - 3} \right)}^2} + {{\left( {16 - \left( { - 2} \right)} \right)}^2}} \\
   = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( {16 + 2} \right)}^2}} \\
   = \sqrt {36 + {{\left( {18} \right)}^2}} \\
   = \sqrt {36 + 324} \\
   = \sqrt {360} \\
   = \sqrt {36 \times 10} \\
   = 6\sqrt {10} units \\
 $
Distance between C and A= $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
\[\left( {x1,y1} \right) = \left( { - 3,16} \right),\left( {x2,y2} \right) = \left( {1,4} \right)\]
CA= $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
$
   = \sqrt {{{\left( {1 - \left( { - 3} \right)} \right)}^2} + {{\left( {4 - 16} \right)}^2}} \\
   = \sqrt {{{\left( 4 \right)}^2} + {{\left( { - 12} \right)}^2}} \\
   = \sqrt {16 + {{\left( {12} \right)}^2}} \\
   = \sqrt {16 + 144} \\
   = \sqrt {160} \\
   = \sqrt {16 \times 10} \\
   = 4\sqrt {10} units \\
 $
$AB = 2\sqrt {10} ,BC = 6\sqrt {10} ,CA = 4\sqrt {10} $
Here, we can clearly notice that AB and CA give BC
$
  2\sqrt {10} + 4\sqrt {10} = 6\sqrt {10} \\
   \to AB + CA = BC \\
 $
Therefore, the points (1, 4), (3, -2) and (-3, 16) are collinear.
b) We are given (1, 1), (t, 3), (4, 6) and (-2, 5) are the vertices of a quadrilateral and the area of the quadrilateral is 13 sq. units.
Area of a quadrilateral when the vertices are given is $ = \dfrac{1}{2}\left\{ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_4} + {x_4}{y_1}} \right) - \left( {{x_2}{y_1} + {x_3}{y_2} + {x_4}{y_3} + {x_1}{y_4}} \right)} \right\}$
$\left( {{x_1},{y_1}} \right) = \left( {1,4} \right),\left( {{x_2},{y_2}} \right) = \left( {t,3} \right),\left( {{x_3},{y_3}} \right) = \left( {4,6} \right),\left( {{x_4},{y_4}} \right) = \left( { - 2,5} \right)$
Area of the given quadrilateral is
$
   = \dfrac{1}{2}\left\{ {\left( {1 \times 3 + t \times 6 + 4 \times 5 + \left( { - 2} \right) \times 4} \right) - \left( {t \times 4 + 4 \times 3 + \left( { - 2} \right) \times 6 + 1 \times 5} \right)} \right\} \\
   = \dfrac{1}{2}\left\{ {\left( {3 + 6t + 20 - 8} \right) - \left( {4t + 12 - 12 + 5} \right)} \right\} \\
   = \dfrac{1}{2}\left\{ {\left( {6t + 15} \right) - \left( {4t + 5} \right)} \right\} \\
   = \dfrac{1}{2}\left\{ {\left( {6t + 15 - 4t - 5} \right)} \right\} \\
   = \dfrac{1}{2}\left\{ {\left( {2t + 10} \right)} \right\} \\
   = \dfrac{1}{2}\left\{ {2\left( {t + 5} \right)} \right\} \\
   = t + 5 \\
 $
Area of the quadrilateral given in the question is 13 sq. units.
$
  t + 5 = 13 \\
  t = 8 \\
$
Therefore, the value of ‘t’ is 8.


Note: A quadrilateral has four sides and it is a two-dimensional shape. Rectangle, square, parallelogram, rhombus etc are some of the examples of quadrilaterals. Sum of the interior angles of a quadrilateral is 360 °. Collinear points are the points that lie on the same line.