Advertisements
Advertisements
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 |
Advertisements
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
APPEARS IN
संबंधित प्रश्न
Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, –4) and B (8, 0).
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.
Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and (–3, 2) are vertices of a parallelogram.
A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k – y1 = m (h – x1).
Consider the given population and year graph. Find the slope of the line AB and using it, find what will be the population in the year 2010?
Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x –2y = 3.
Find the slope of the lines which make the following angle with the positive direction of x-axis:
\[- \frac{\pi}{4}\]
State whether the two lines in each of the following is parallel, perpendicular or neither.
Through (6, 3) and (1, 1); through (−2, 5) and (2, −5)
Find the slope of a line (i) which bisects the first quadrant angle (ii) which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
What is the value of y so that the line through (3, y) and (2, 7) is parallel to the line through (−1, 4) and (0, 6)?
Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2) and C (3, 3) are the vertices of a right angled triangle.
Find the angle between the X-axis and the line joining the points (3, −1) and (4, −2).
Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear.
Find the equation of a straight line with slope − 1/3 and y-intercept − 4.
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Find the angles between the following pair of straight lines:
3x − y + 5 = 0 and x − 3y + 1 = 0
Find the acute angle between the lines 2x − y + 3 = 0 and x + y + 2 = 0.
Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.
If θ is the angle which the straight line joining the points (x1, y1) and (x2, y2) subtends at the origin, prove that \[\tan \theta = \frac{x_2 y_1 - x_1 y_2}{x_1 x_2 + y_1 y_2}\text { and } \cos \theta = \frac{x_1 x_2 + y_1 y_2}{\sqrt{{x_1}^2 + {y_1}^2}\sqrt{{x_2}^2 + {y_2}^2}}\].
Find the tangent of the angle between the lines which have intercepts 3, 4 and 1, 8 on the axes respectively.
The acute angle between the medians drawn from the acute angles of a right angled isosceles triangle is
The reflection of the point (4, −13) about the line 5x + y + 6 = 0 is
If the slopes of the lines given by the equation ax2 + 2hxy + by2 = 0 are in the ratio 5 : 3, then the ratio h2 : ab = ______.
The equation of a line passing through the point (7, - 4) and perpendicular to the line passing through the points (2, 3) and (1 , - 2 ) is ______.
If x + y = k is normal to y2 = 12x, then k is ______.
Point of the curve y2 = 3(x – 2) at which the normal is parallel to the line 2y + 4x + 5 = 0 is ______.
If the slope of a line passing through the point A(3, 2) is `3/4`, then find points on the line which are 5 units away from the point A.
If one diagonal of a square is along the line 8x – 15y = 0 and one of its vertex is at (1, 2), then find the equation of sides of the square passing through this vertex.
The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 are ______.
Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, – 1).
Find the angle between the lines y = `(2 - sqrt(3)) (x + 5)` and y = `(2 + sqrt(3))(x - 7)`
Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are ______.
One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is ______.
If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.
The line `x/a + y/b` = 1 moves in such a way that `1/a^2 + 1/b^2 = 1/c^2`, where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is x2 + y2 = c2.
The line which passes through the origin and intersect the two lines `(x - 1)/2 = (y + 3)/4 = (z - 5)/3, (x - 4)/2 = (y + 3)/3 = (z - 14)/4`, is ______.
If the line joining two points A (2, 0) and B (3, 1) is rotated about A in anticlockwise direction through an angle of 15°, then the equation of the line in new position is ______.
