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प्रश्न
Show that points P(1, –2), Q(5, 2), R(3, –1), S(–1, –5) are the vertices of a parallelogram.
उत्तर
Let, P(1, −2) = (x1, y1); Q(5, 2) = (x2, y2); R(3, −1) = (x3, y3) and S(−1, −5) = (x4, y4).
By distance formula,
PQ = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
= `sqrt((5 - 1)^2 + [2 - (-2)]^2)`
= `sqrt((4)^2 + (4)^2)`
= `sqrt(16 + 16)`
= `sqrt32`
= `sqrt(2 xx 2 xx 2 xx 2 xx 2)`
= `4sqrt2` ...(1)
QR = `sqrt((x_3 - x_2)^2 + (y_3 - y_2)^2)`
= `sqrt((3 - 5)^2 + (-1 - 2)^2`
= `sqrt((-2)^2 + (-3)^2)`
= `sqrt(4 + 9)`
= `sqrt13` ...(2)
RS = `sqrt((x_4 - x_3)^2 + (y_4 - y_3)^2)`
= `sqrt((-1 - 3)^2 + [-5 - (-1)]^2)`
= `sqrt((-4)^2 + (-4)^2)`
= `sqrt(16 + 16)`
= `sqrt32`
= `sqrt(2 xx 2 xx 2 xx 2 xx 2)`
= `4sqrt2` ...(3)
PS = `sqrt((x_4 - x_1)^2 + (y_4 - y_1)^2)`
= `sqrt((-1 - 1)^2 + [-5 - (-2)]^2)`
= `sqrt((-2)^2 + (-5 + 2)^2)`
= `sqrt((-2)^2 + (-3)^2)`
= `sqrt(4 + 9)`
= `sqrt13` ...(4)
In `square`PQRS,
PQ = RS ...[From (1) and (3)]
QR = PS ...[From (2) and (4)]
`square`PQRS is a parallelogram. ...(A quadrilateral is a parallelogram if both the pairs of its opposite sides are congruent.)
Points P(1, –2), Q(5, 2), R(3, –1), S(–1, –5) are the vertices of a parallelogram.
संबंधित प्रश्न
Find the slope of the line with inclination 30° .
Find the slope of the line passing through the points A(-2, 1) and B(0, 3).
Find the slope of the line passing through the points A(2, 3) and B(4, 7).
Find the slope of the line parallel to AB if : A = (−2, 4) and B = (0, 6)
Find the slope of the line parallel to AB if : A = (0, −3) and B = (−2, 5)
(−2, 4), (4, 8), (10, 7) and (11, –5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.
Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.
If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
Angle made by the line with the positive direction of X-axis is given. Find the slope of the line.
45°
Show that A(–4, –7), B (–1, 2), C (8, 5) and D (5, –4) are the vertices of a parallelogram.
Find the slope of a line, correct of two decimals, whose inclination is 75°
Find the slope and the y-intercept of the following line 4y = 5x - 8
Find slope of a line passing through the points A(3, 1) and B(5, 3).
The line through A (- 2, 3) and B (4, b) is perpendicular to the line 2a – 4y = 5. Find the value of b.
Show that the line joining (2, – 3) and (- 5, 1) is:
(i) Parallel to line joining (7, -1) and (0, 3).
(ii) Perpendicular to the line joining (4, 5) and (0, -2).
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.
If the lines 7y = ax + 4 and 2y = 3 − x, are parallel to each other, then the value of ‘a’ is:
