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प्रश्न
| Column C1 | Column C2 |
| (a) The coordinates of the points P and Q on the line x + 5y = 13 which are at a distance of 2 units from the line 12x – 5y + 26 = 0 are |
(i) (3, 1), (–7, 11) |
| (b) The coordinates of the point on the line x + y = 4, which are at a unit distance from the line 4x + 3y – 10 = 0 are |
(ii) `(- 1/3, 11/3), (4/3, 7/3)` |
| (c) The coordinates of the point on the line joining A (–2, 5) and B (3, 1) such that AP = PQ = QB are |
(iii) `(1, 12/5), (-3, 16/5)` |
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उत्तर
| Column C1 | Column C2 |
| (a) The coordinates of the points P and Q on the line x + 5y = 13 which are at a distance of 2 units from the line 12x – 5y + 26 = 0 are |
(i) `(1, 12/5), (-3, 16/5)` |
| (b) The coordinates of the point on the line x + y = 4, which are at a unit distance from the line 4x + 3y – 10 = 0 are |
(ii) (3, 1), (–7, 11) |
| (c) The coordinates of the point on the line joining A (–2, 5) and B (3, 1) such that AP = PQ = QB are |
(iii) `(- 1/3, 11/3), (4/3, 7/3)` |
Explanation:
(a) Let P(x1, y1)be any point on the given line x + 5y = 13
∴ x1 + 5y1 = 13
Distance of line 12x – 5y + 26 = 0 from the point P(x1, y1)
2 = `|(12x_1 - 5y_1 + 26)/sqrt((12)^2 + (-5)^2)|`
⇒ 2 = `|(12x_1 - (13 - x_1) + 26)/13|`
⇒ 2 = `|(12x_1 - 13 + x_1 + 26)/13|`
⇒ 2 = `|(13x_1 + 13)/13|`
⇒ 2 = ± (x1 + 1)
⇒ 2 = x1 + 1
⇒ x1 = 1 ......(Taking (+) sign)
And 2 = – x1 – 1
⇒ x1 = – 3 ......(Taking (–) sign)
Putting the values of x1 in equation x1 + 5y1 = 13.
We get y1 = `12/5` and `16/5`.
So, the required points are `(1, 12/5)` and `(-3, 16/5)`.
(b) Let P(x1, y1) be any point on the given line x + y = 4
∴ x1 + y1 = 4 ......(i)
Distance of the line 4x + 3y – 10 = 0 from the point P(x1, y1)
1 = `|(4x_1 + 3y_1 - 10)/sqrt((4)^2 + (3)^2)|`
⇒ 1 = `|(4x_1 + 3(4 - x_1) - 10)/5|`
⇒ 1 = `|(4x_1 + 12 - 3x_1 - 10)/5|`
⇒ 1 = `|(x_1 + 2)/5|`
⇒ 1 = `+- ((x_1 + 2)/5)`
⇒ `(x_1 + 2)/5` = 1 ......(Taking (+) sign)
⇒ x1 + 2 = 5
⇒ x1 = 3
And `(x_1 + 2)/5 = - 1` ......(Taking (–) sign)
⇒ x1 + 2 = 5
⇒ x1 = 3
Putting the values of x1 in equation (i) we get
x1 + y1 = 4
At x1 = 3, y1 = 1
At x1 = – 7, y1 = 11
So, the required points are (3, 1) and (– 7, 11).
(c) Given that AP = PQ = QB
Equation of line joining A(– 2, 5) and B(3, 1) is
y – 5 = `(1 - 5)/(3 + 2) (x + 2)`
⇒ y – 5 = `(-4)/5 (x + 2)`
⇒ 5y – 25 = – 4x – 8
⇒ 4x + 5y – 17 = 0
Let P(x1, y1) and Q(x2, y2) be any two points on the line AB
P(x1, y1) divides the line AB in the ratio 1 : 2
∴ x1 = `(1.3 + 2(-2))/(1 + 2)`
= `(3 - 4)/3`
= `(-1)/3`
y1 = `(1.1 + 2.5)/(1 + 2)`
= `(1 + 10)3`
= `11/3`
So, the coordinates of P(x1, y1) = `((-1)/3, 11/3)`
Now point Q(x2, y2) is the mid-point of PB
∴ x2 = `(3 - 1/3)/2 = 4/3`
y2 = `(1 + 11/3)/2 = 7/3`
Hence, the coordinates of Q(x2, y2) = `(4/3, 7/3)`
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| Column C1 | Column C2 |
| (a) Through the point (2, 1) is | (i) 2x – y = 4 |
| (b) Perpendicular to the line (ii) x + y – 5 = 0 x + 2y + 1 = 0 is |
(ii) x + y – 5 = 0 |
| (c) Parallel to the line (iii) x – y –1 = 0 3x – 4y + 5 = 0 is |
(iii) x – y –1 = 0 |
| (d) Equally inclined to the axes is | (iv) 3x – 4y – 1 = 0 |
