Question

The area of the triangle whose two of the vertices are A(0, 1), B(3, 4) and the intersection point of median from A with side BC is (–2, 3), is (in square units)

Answer 1

Answer:

Given: The area of a parallelogram is 12. A(-1, 3) and B(-2, 4) To find: Find the other two vertices of the parallelogram. Let C is (x, y) and A(-1, 3) Since, AC is bisected at P, y coordinate ( when p = 0) Then, (y + 3)/2 = 0 y = -3 So, Coordinate of C is (x, -3) Now, Area of parallelogram ABCD = area of triangle ABC + Area of triangle BAD Since, 2(Area of triangles) = area of parallelogram We have, A(-1, 3) B(-2, 4) and C(x, -3) 12 = 5 - x So, x = -7 Hence, Coordinate of C is (-7, -3) In the same we will calculate for D Let D is (x, y) and A(-2, 4) Since, BD is bisected at Q, y coordinate ( when Q = 0) Then, (y + 4)/2 y = -4 So, Coordinate of C is (x, -4) We have, A(-1, 3) B(-2, 4) and C(x, -4) 12 = 6 - x So, x = -6 Hence, Coordinate of D is (-6, -4) Hence, C (-7, -3) and D(-6, -4)Read more on Sarthaks.com - https://www.sarthaks.com/963288/the-area-of-a-parallelogram-is-12-square-units-two-of-its-vertices-are-the-points-a-1-3-and-b-2-4?show=963288#q963288

Answer 2

The answer is 12 square units

Given: Two vertices of triangle A(0, 1), B(3, 4)

The intersection of point from A with side BC is (-2, 3)

To find: Area of triangle

Solution: To find area of triangle we need to know the 3rd vertice of triangle, Let C(x, y) be the 3rd vertices of triangle

Given the point of intersection of median from A to BC is (–2, 3)

Here the point of intersection of median will divide the BC equal

then the point of intersection = midpoint of BC

⇒ Midpoint of B(3, 4) and C(x, y) = (-2, 3)  

⇒ $(\frac{3+x}{2}, \frac{4+y}{2} ) = (-2, 3)$    

⇒  $\frac{3+x}{2} = -2$               ⇒ $\frac{4+y}{2} = 3$

⇒ 3 + x = - 4               ⇒ 4 + y = 6

⇒ x = -7                       ⇒ y = 2  

Therefore, 3rd vertice of triangle = (-1, 2)

Now find area of triangle with A(0, 1), B(3, 4) and C(-7, 2)  

We know area of triangle = $\frac{1}{2} | x_{1} (y_{2} - y_{3} ) + x_{2} (y_{3} -y_{1} ) + x_{3} (y_{1} -y_{2})$  

= $\frac{1}{2} | 0 (4- 2 ) +3 (2 -1 ) + (-7) (1 -4)|$

=  $\frac{1}{2} |0 +3 (1 ) -7 (-3)|$

= $\frac{1}{2} |0 +3 (1 ) 21|$

=  $\frac{1}{2} |24|$ = 12 square units

Area of triangle = 12 square units

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