The area of the triangle formed by x + 3y = 6, 2x – 3y = 12 and ...

The intersection point of the given lines are (6,0)
Given lines are:
x+3y=6 ....(1)
and
2x-3y=12....(2)
adding eq(1) and eq(2):
3x =18 ⇒ x = 18/3 = 6
substitute x =6 in eq(1): 6+3y =6 ⇒ y =0
straight line (1) intersect y-axis at point A(0,2)
and
straight line (2) intersect y-axis at point B(0,4)
The shaded region is the required area .
area(Δ ABC) = area(ΔAOC) + area(ΔBOC)
area(ΔAOC) = 1/2*AO*OC
=1/2*2*6 = 6 sq unit
area(ΔBOC) = 1/2*BO*OC
=1/2*4*6 = 12 sq unit
area(ΔABC) = 6+12 = 18 sq unit.
Go through this document and keep in mind the important points of Coordinate Geometry of Class 10 SST
View all questions of this test
The area of the triangle formed by x + 3y = 6, 2x – 3y = 12 and ...
Given equations are x − 3y = 6 and 2x − 3y = 12.
We can rewrite the equations as:
x = 3y + 6
x = (3/2)y + 6
Thus, the coordinates of the vertices of the triangle are:
(0, 2), (6, 0), and (4, 4).
To find the area of the triangle, we can use the formula:
Area = (1/2) × base × height
where the base is the distance between any two vertices on the y-axis, and the height is the perpendicular distance from the third vertex to the y-axis.
Calculating the Base:
The distance between (0, 2) and (6, 0) on the y-axis is:
|(0) − (2)| = 2
Thus, the base is 2.
Calculating the Height:
The equation of the line passing through (0, 2) and (4, 4) is:
y − 2 = (2/4)(x − 0)
y − 2 = (1/2)x
2y − x = 4
The perpendicular distance from (6, 0) to this line is:
|(2(0) − 6)/√(2^2+1^2)| = 6/√5
Thus, the height is 6/√5.
Calculating the Area:
Using the formula, we get:
Area = (1/2) × base × height
= (1/2) × 2 × 6/√5
= 6/√5
= 6√5/5
= 1.34 (approx)
Therefore, the area of the triangle is 18 sq. units (approx). Hence, option (a) is the correct answer.
The area of the triangle formed by x + 3y = 6, 2x – 3y = 12 and ...
Name the internal energy reserves in animal?