Advertisements
Advertisements
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
Advertisements
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
APPEARS IN
संबंधित प्रश्न
Draw the graph of the equation given below.
3x - y = 0
Draw the graph for the linear equation given below:
x = 3
Draw the graph for the linear equation given below:
y = - 2x
Draw the graph for the equation given below:
`(2x - 1)/(3) - (y - 2)/(5) = 0`
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
Draw the graph of equation x + 2y - 3 = 0. From the graph, find:
(i) x1, the value of x, when y = 3
(ii) x2, the value of x, when y = - 2.
Draw a graph of each of the following equations: y = `(5)/(2) xx + (2)/(5)`
Draw a graph of each of the following equations: y = `(3)/(5) x - 1`
Draw a graph of the equation 2x + 3y + 5 = 0, from the graph find the value of:
(i) x, when y = -3
(ii) y, when x = 8
Draw a graph of the equation 5x - 3y = 1. From the graph find the value of:
(i) x, when y = 8
(ii) y, when x = 2
