Advertisements
Advertisements
Question
Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(−2, 6) and C(3, 1) is 10 square units.
Advertisements
Solution
The formula for the area ‘A’ encompassed by three points( x1 , y1) , (x2 , y2) and (x3 , y3) is given by the formula,
\[∆ = \frac{1}{2}\left| \left( x_1 y_2 + x_2 y_3 + x_3 y_1 \right) - \left( x_2 y_1 + x_3 y_2 + x_1 y_3 \right) \right|\]
The three given points are A(a,2a), B(−2,6) and C(3,1). It is also said that the area enclosed by them is 10 square units. Substituting these values in the above mentioned formula we have,
\[∆ = \frac{1}{2}\left| \left( a \times 6 + \left( - 2 \right) \times 1 + 3 \times 2a \right) - \left( \left( - 2 \right) \times 2a + 3 \times 6 + a \times 1 \right) \right|\]
\[10 = \frac{1}{2}\left| \left( 6a - 2 + 6a \right) - \left( - 4a + 18 + a \right) \right|\]
\[10 = \frac{1}{2}\left| 15a - 20 \right|\]
\[20 = \left| 15a - 20 \right|\]
\[4 = \left| 3a - 4 \right|\]
We have |3a - 4 | = 4. Hence either
3a - 4 = 4
3a = 8
a = `8/3`
Or
-(3a - 4 ) = 4
4 - 3a = 4
a = 0
Hence the values of ‘a’ which satisfies the given conditions are a = 0 `a = 8/3` .
APPEARS IN
RELATED QUESTIONS
On which axis do the following points lie?
R(−4,0)
Prove that the points (−2, 5), (0, 1) and (2, −3) are collinear.
Which point on the y-axis is equidistant from (2, 3) and (−4, 1)?
Three consecutive vertices of a parallelogram are (-2,-1), (1, 0) and (4, 3). Find the fourth vertex.
If three consecutive vertices of a parallelogram are (1, -2), (3, 6) and (5, 10), find its fourth vertex.
Show that the points A (1, 0), B (5, 3), C (2, 7) and D (−2, 4) are the vertices of a parallelogram.
If p(x , y) is point equidistant from the points A(6, -1) and B(2,3) A , show that x – y = 3
Show hat A(1,2), B(4,3),C(6,6) and D(3,5) are the vertices of a parallelogram. Show that ABCD is not rectangle.
Find the ratio in which the point (−3, k) divides the line-segment joining the points (−5, −4) and (−2, 3). Also find the value of k ?
Find the area of the quadrilateral ABCD, whose vertices are A(−3, −1), B (−2, −4), C(4, − 1) and D (3, 4).
The area of the triangle formed by the points P (0, 1), Q (0, 5) and R (3, 4) is
If (a,b) is the mid-point of the line segment joining the points A (10, - 6) , B (k,4) and a - 2b = 18 , find the value of k and the distance AB.
If the point \[C \left( - 1, 2 \right)\] divides internally the line segment joining the points A (2, 5) and B( x, y ) in the ratio 3 : 4 , find the value of x2 + y2 .
Find the area of a parallelogram ABCD if three of its vertices are A(2, 4), B(2 + \[\sqrt{3}\] , 5) and C(2, 6).
If (x , 2), (−3, −4) and (7, −5) are collinear, then x =
The line segment joining points (−3, −4), and (1, −2) is divided by y-axis in the ratio.
If the line segment joining the points (3, −4), and (1, 2) is trisected at points P (a, −2) and Q \[\left( \frac{5}{3}, b \right)\] , Then,
Write the equations of the x-axis and y-axis.
Which of the points P(0, 3), Q(1, 0), R(0, –1), S(–5, 0), T(1, 2) do not lie on the x-axis?
Point (3, 0) lies in the first quadrant.
