Advertisements
Advertisements
प्रश्न
In the following, find the coordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation:
3 - 2x = 7; 2y + 1 = 10 - 2`(1)/(2)`y.
उत्तर
3 - 2x = 7; 2y + 1 = 10 - 2`(1)/(2)`y
Now
3 - 2x = 7
3 - 7 = 2x
-4 = 2x
-2 = x
Again
2y + 1 = 10 - 2`(1)/(2)`y
2y + 1 = 10 -`(5)/(2)`y
4y + 2 = 20 - 5y
4y + 5y = 20 - 2
9y = 18
y = 2
∴ The co-ordinates of the point (-2, 2)
APPEARS IN
संबंधित प्रश्न
Find the values of x and y if:
(x - 1, y + 3) = (4, 4)
State, true or false:
The origin (0, 0) lies on the x-axis.
In the following, find the co-ordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation:
`5x - (5 - x) = (1)/(2) (3 - x); 4 -3y = (4 + y)/(3)`
Find the co-ordinates of points whose: Abscissa is 6 and ordinate is 2
Find the co-ordinates of points whose: Abscissa is 5 and ordinate is -1
Find the co-ordinates of points whose: Abscissa is -2 and ordinate is 0
Find the co-ordinates of points whose: Abscissa is 0 and ordinate is 0
Find the co-ordinates of points whose: Abscissa is -7 and ordinate is 4
In each of the following, find the coordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation.
5 + 2x = `9: 3(1)/(2)y + 1` = 12 - 3y
A rectangle PQRS is drawn on the coordinate axes such that P is the origin, PQ = 6 units and PS = 5 units. Find the coordinates of the vertices P, Q R and S. Also, find the area of the rectangle.
