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प्रश्न
For the linear equation, given above, draw the graph and then use the graph drawn (in the following case) to find the area of a triangle enclosed by the graph and the co-ordinates axes:
7 - 3 (1 - y) = -5 + 2x
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उत्तर
First draw the graph as follows:
This is a right triangle.
Thus the area of the triangle will be:
A = `(1)/(2) xx "base" xx "altitude"`
= `(1)/(2) xx (9)/(2) xx 3`
= `(27)/(4)`
= 6.75 sq.units
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संबंधित प्रश्न
Draw the graph of the equation given below.
3x - y = 0
Draw the graph for the linear equation given below:
x + 3 = 0
Draw the graph for the linear equation given below:
2x - 7 = 0
Draw the graph for the linear equation given below:
x - 3 = `(2)/(5)(y + 1)`
For the linear equation, given above, draw the graph and then use the graph drawn (in the following case) to find the area of a triangle enclosed by the graph and the co-ordinates axes:
3x − (5 − y) = 7
For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.
y = 3x - 1
y = 3x + 2
On the same graph paper, plot the graph of y = x - 2, y = 2x + 1 and y = 4 from x= - 4 to 3.
Use the graphical method to show that the straight lines given by the equations x + y = 2, x - 2y = 5 and `x/(3) + y = 0` pass through the same point.
Draw a graph of each of the following equations: x + 6y = 15
Draw a graph of each of the following equations: `(x - 2)/(3) - (y + 1)/(2)` = 0
