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प्रश्न
The line through P (5, 3) intersects Y axis at Q.
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the coordinates of Q.
उत्तर
(i)
m = tan θ = tan 45°
m = 1
(ii) Equation of line PQ
y - y1 = m(x - x1)
y - 3 = 1 (x - 5)
y - 3 = x - 5
⇒ x - y - 2 = 0
(iii) Equation of PQ is
x - y - 2 = 0
Put x = 0 (coordinates of Q)
-y -2 = 0
⇒ y = -2
So, coordinates of Q(0, -2).
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संबंधित प्रश्न
Find the slope of the line parallel to AB if : A = (0, −3) and B = (−2, 5)
Find the slope of the line perpendicular to AB if : A = (0, −5) and B = (−2, 4)
The line passing through (0, 2) and (−3, −1) is parallel to the line passing through (−1, 5) and (4, a). Find a.
Without using the distance formula, show that the points A(4, 5), B(1, 2), C(4, 3) and D(7, 6) are the vertices of a parallelogram.
Find the value(s) of k so that PQ will be parallel to RS. Given : P(5, −1), Q(6, 11), R(6, −4k) and S(7, k2)
Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.
Find the slope of the line passing through the points G(4, 5) and H (–1, –2).
Show that the line joining the points A(4, 8) and B(5, 5) is parallel to the line joining the points C(2, 4) and D(1, 7).
Find the slope of the line passing through the points M(4,0) and N(-2,-3).
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.
