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प्रश्न
(1, 5) and (–3, –1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.
उत्तर
A = (1, 5) and C = (–3, –1)
We know that in a rhombus, diagonals bisect each other at right angle.
Let O be the point of intersection of the diagonals AC and BD.
Co-ordinates of O are
`((1 - 3)/2, (5 - 1)/2) = (-1, 2)`
Slope of AC = `(-1 - 5)/(-3 - 1) = (-6)/-4 = 3/2`
For line AC:
Slope = m = `3/2, (x_1, y_1) = (1, 5)`
Equation of the line AC is
y – y1 = m(x – x1)
`y - 5 = 3/2 (x - 1)`
2y − 10 = 3x − 3
3x − 2y + 7 = 0
For line BD:
Slope = m = `(-1)/("slope of AC") = (-2)/3`,
(x1, y1) = (−1, 2)
Equation of the line BD is
y – y1 = m(x – x1)
`y - 2 = (-2)/3 (x + 1)`
3y − 6 = −2x − 2
2x + 3y = 4
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संबंधित प्रश्न
Use graph paper for this question (Take 2 cm = 1 unit along both x and y-axis). ABCD is a quadrilateral whose vertices are A(2, 2), B(2, –2), C(0, –1) and D(0, 1).
1) Reflect quadrilateral ABCD on the y-axis and name it as A'B'CD
2) Write down the coordinates of A' and B'.
3) Name two points which are invariant under the above reflection
4) Name the polygon A'B'CD
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1) the coordinate of the fourth vertex D
2) length of diagonal BD
3) equation of the side AD of the parallelogram ABCD
Using a graph paper, plot the points A(6, 4) and B(0, 4).
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- Write the co-ordinates of A' and B'.
- State the geometrical name for the figure ABA'B'.
- Find its perimeter.
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- Find the co-ordinates of its fourth vertex D, if ABCD is a square.
- Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
O(0, 0), A(3, 5) and B(−5, −3) are the vertices of triangle OAB. Find the equation of median of triangle OAB through vertex O.
O(0, 0), A(3, 5) and B(−5, −3) are the vertices of triangle OAB. Find the equation of altitude of triangle OAB through vertex B.
A line AB meets the x-axis at point A and y-axis at point B. The point P(−4, −2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:
- the co-ordinates of A and B.
- equation of line through P and perpendicular to AB.
Point A and B have co-ordinates (7, −3) and (1, 9) respectively. Find:
- the slope of AB.
- the equation of perpendicular bisector of the line segment AB.
- the value of ‘p’ of (−2, p) lies on it.
Use a graph sheet for this question.
Take 1 cm = 1 unit along both x and y axis.
(i) Plot the following points:
A(0,5), B(3,0), C(1,0) and D(1,–5)
(ii) Reflect the points B, C and D on the y axis and name them as B',C'andD' respectively.
(iii) Write down the coordinates of B',C 'and D'
(iv) Join the point A, B, C, D, D ', C ', B', A in order and give a name to the closed figure ABCDD'C'B
A line is of length 10 units and one end is at the point (2, – 3). If the abscissa of the other end be 10, prove that its ordinate must be 3 or – 9.
