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प्रश्न
The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and
| 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 |
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उत्तर
| Column C1 | Column C2 |
| (a) Through the point (2, 1) is | (i) x – y –1 = 0 |
| (b) Perpendicular to the line (ii) x + y – 5 = 0 x + 2y + 1 = 0 is |
(ii) 2x – y = 4 |
| (c) Parallel to the line (iii) x – y –1 = 0 3x – 4y + 5 = 0 is |
(iii) 3x – 4y – 1 = 0 |
| (d) Equally inclined to the axes is | (iv) x + y – 5 = 0 |
Explanation:
(a) Given equations are 2x – 3y = 0 ......(i)
And 4x – 5y = 2 ......(ii)
Equations of line passing through eq. (i) and (ii) we get
(2x – 3y) + k(4x – 5y –2) = 0 .....(iii)
If equation (iii) passes through (2, 1), we get
(2 × 2 – 3 × 1) + k(4 × 2 – 5 × 1 – 2) = 0
⇒ (4 – 3) + k(8 – 5 – 2) = 0
⇒ 1 + k(8 – 7) = 0
⇒ k = – 1
So, the required equation is
(2x – 3y) – 1(4x – 5y – 2) = 0
⇒ 2x – 3y – 4x + 5y + 2 = 0
⇒ – 2x + 2y + 2 = 0
⇒ x – y – 1 = 0
(b) Equation of any line passing through the point of intersection of the line 2x – 3y = 0 and 4x – 5y = 2 is
(2x – 3y) + k(4x – 5y – 2) = 0 ......(i)
⇒ (2 + 4k)x + (– 3 – 5k)y – 2k = 0
Slope = `(-(2 + 4k))/(-3 - 5k) = (2 + 4k)/(3 + 5k)`
Slope of the given line x + 2y + 1 = 0 is `- 1/2`.
If they are perpendicular to each other then
`- 1/2((2 + 4k)/(3 + 5k)) = -1`
⇒ `(1 + 2k)/(3 + 5k)` = 1
⇒ 1 + 2k = 3 + 5k
⇒ 3k = – 2
⇒ k = `(-2)/3`
Putting the value of k is eq. (i) we get
`(2x - 3y) - 2/3 (4x - 5y - 2)` = 0
⇒ 6x – 9y – 8x + 10y + 4 = 0
⇒ – 2x + y + 4 = 0
⇒ 2x – y = 4
(c) Given equations are
2x – 3y = 0 .......(i)
4x – 5y = 2 ......(ii)
Equation of line passing through equation (i) and (ii) we get
(2x – 3y) + k(4x – 5y – 2) = 0
⇒ (2 + 4k)x + (– 3 – 5k)y – 2k = 0
Slope = `(-(2 + 4k))/(-3 - 5k) = (2 + 4k)/(3 + 5k)`
Slope of the given line 3x – 4y + 5 = 0 is `3/4`.
If the two equations are parallel, then
`(2 + 4k)/(3 + 5k) = 3/4`
⇒ 8 + 16k = 9 + 15k
⇒ 16k – 15k = 9 – 8
⇒ k = 1
So, the equation of the required line is
(2x – 3y) + 1(4x – 5y – 2) = 0
2x – 3y + 4x – 5y – 2 = 0
⇒ 6x – 8y – 2 = 0
⇒ 3x – 4y – 1 = 0
(d) Given equations are
2x – 3y = 0 ......(i)
4x – 5y – 2 = 0 ......(ii)
Equation of line passing through the intersection of equation (i) and (ii) we get
(2x – 3y) + k(4x – 5y – 2) = 0
⇒ (2 + 4k)x + (– 3 – 5k)y – 2k = 0
Slope = `(2 + 4k)/(3 + 5k)`
Since the equation is equally inclined with axes
∴ Slope = tan 135° = tan(180° – 45°)
= – tan 45° = – 1
So `(2 + 4k)/(3 + 5k) = -1`
⇒ 2 + 4k = – 3 – 5k
⇒ 4k + 5k = – 3 – 2
⇒ 9k = – 5
⇒ k = `(-5)/9`
Required equation is `(2x - 3y) - 5/9 (4x - 5y - 2)` = 0
⇒ 18x – 27y – 20x + 25y + 10 = 0
⇒ – 2x – 2y + 10 = 0
⇒ x + y – 5 = 0
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