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Question
A line passing through the points (a, 2a) and (- 2, 3) is perpendicular to the line 4a + 3y + 5 = 0. Find the value of a.
Solution
Let m1 be the slope of the joining at the points (a, 2a) and (-2, 3), then
m1 = `(2a - 3)/(a + 2)`
Also slope of the line 4x + 3y + 5 = 0.
m2 = `-(4)/(3)`
Since, both the line are perpendicular.
So, m1m2 = -1
⇒ `(2a - 3)/(a + 2) xx ((-4))/(3)` = -1
⇒ 8a - 12 = 3a + 6
⇒ 8a - 3a = 18
⇒ 5a = 18
⇒ a = `(18)/(5)`
⇒ a = `3(3)/(5)`.
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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.
