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प्रश्न
Prove that A(4, 3), B(6, 4), C(5, 6) and D(3, 5) are the angular points of a square.
उत्तर
Now
AB = `sqrt((4 - 6)^2 + (3 - 4)^2)`
= `sqrt(4 + 1) = sqrt(5)"units"`
BC = `sqrt((6 - 5)^2 + (4 - 6)^2) = sqrt(1 + 4)`
BC = `sqrt(5)"units"`.
CD = `sqrt((5 - 3)^2 + (6 - 5)^2) = sqrt(4 + 1)`
CD = `sqrt(5)"units"`
Also DA = `sqrt((4 - 3)^2 + (3 + 5)^2)`
= `sqrt(1 + 4) = sqrt(5)`
DA = `sqrt(5)"units"`
So AB = BC = CD = DA.
Now slope of AB = m1 - `(4 - 3)/(6 - 4) = (1)/(2)`
Slope of BC = m2 = `(6 - 4)/(5 - 6) = (2)/(-1)`
Slope of CA = m3 = `(5 - 6)/(3 - 5) = (1)/(2)`
Slope of DA = m4 = `(5 - 3)/(3 - 4) = (2)/(-1)`
Since m1 = m3 and = m2 = m4
So AB || CD
and BC || DA.
Therefore, AB ⊥ BC
∴ ABCD is a square.
Hence proved.
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संबंधित प्रश्न
Find the slope of the line parallel to AB if : A = (0, −3) and B = (−2, 5)
Without using the distance formula, show that the points A(4, −2), B(−4, 4) and C(10, 6) are the vertices of a right-angled triangle.
Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
Find the slope of the lines passing through the given point.
A(2, 3), B(4, 7)
Find k, if PQ || RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k).
Find the slope of a line passing through the given pair of points (3,7) and (5,13)
Find the slope of a line parallel to the given line x +3y = 7
Show that the points A(- 2, 5), B(2, – 3) and C(0, 1) are collinear.
If A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of `square`ABCD, show that `square`ABCD is a parallelogram.
Solution:
Slope of line = `("y"_2 - "y"_1)/("x"_2 - "x"_1)`
∴ Slope of line AB = `(2 - 1)/(8 - 6) = square` .......(i)
∴ Slope of line BC = `(4 - 2)/(9 - 8) = square` .....(ii)
∴ Slope of line CD = `(3 - 4)/(7 - 9) = square` .....(iii)
∴ Slope of line DA = `(3 - 1)/(7 - 6) = square` .....(iv)
∴ Slope of line AB = `square` ......[From (i) and (iii)]
∴ line AB || line CD
∴ Slope of line BC = `square` ......[From (ii) and (iv)]
∴ line BC || line DA
Both the pairs of opposite sides of the quadrilateral are parallel.
∴ `square`ABCD is a parallelogram.
Determine whether the following points are collinear. A(–1, –1), B(0, 1), C(1, 3)
Given: Points A(–1, –1), B(0, 1) and C(1, 3)
Slope of line AB = `(square - square)/(square - square) = square/square` = 2
Slope of line BC = `(square - square)/(square - square) = square/square` = 2
Slope of line AB = Slope of line BC and B is the common point.
∴ Points A, B and C are collinear.
