Advertisements
Advertisements
प्रश्न
Find the equation of the perpendicular to the line segment joining (4, 3) and (−1, 1) if it cuts off an intercept −3 from y-axis.
Advertisements
उत्तर
Let m be the slope of the required line.
Here, c = y-intercept = \[-\] 3
Slope of the line joining the points (4, 3) and (−1, 1) = \[\frac{1 - 3}{- 1 - 4} = \frac{2}{5}\]
It is given that the required line is perpendicular to the line joining the points (4, 3) and (−1, 1).
\[\therefore m \times \text { Slope of the line joining the points } \left( 4, 3 \right) and \left( - 1, 1 \right) = - 1\]
\[ \Rightarrow m \times \frac{2}{5} = - 1\]
\[ \Rightarrow m = \frac{- 5}{2}\]
Substituting the values of m and c in y = mx + c, we get:
\[y = - \frac{5}{2}x - 3 \]
\[ \Rightarrow 5 x + 2y + 6 = 0\]
Hence, the equation of the required line is 5x + 2y + 6 = 0.
APPEARS IN
संबंधित प्रश्न
Find the distance between P (x1, y1) and Q (x2, y2) when :
- PQ is parallel to the y-axis,
- PQ is parallel to the x-axis
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).
A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k – y1 = m (h – x1).
Find the values of k for which the line (k–3) x – (4 – k2) y + k2 –7k + 6 = 0 is
- Parallel to the x-axis,
- Parallel to the y-axis,
- Passing through the origin.
Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.
Find the slope of the lines which make the following angle with the positive direction of x-axis:
\[- \frac{\pi}{4}\]
Find the slope of the lines which make the following angle with the positive direction of x-axis: \[\frac{\pi}{3}\]
State whether the two lines in each of the following are parallel, perpendicular or neither.
Through (5, 6) and (2, 3); through (9, −2) and (6, −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.
Using the method of slope, show that the following points are collinear A (16, − 18), B (3, −6), C (−10, 6) .
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)?
What can be said regarding a line if its slope is positive ?
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.
The slope of a line is double of the slope of another line. If tangents of the angle between them is \[\frac{1}{3}\],find the slopes of the other line.
Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear.
By using the concept of slope, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.
Find the equations of the bisectors of the angles between the coordinate axes.
Find the equation of a line which is perpendicular to the line joining (4, 2) and (3, 5) and cuts off an intercept of length 3 on y-axis.
Show that the perpendicular bisectors of the sides of a triangle are concurrent.
If the image of the point (2, 1) with respect to a line mirror is (5, 2), find the equation of the mirror.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (−1, 2).
Find the acute angle between the lines 2x − y + 3 = 0 and x + y + 2 = 0.
Prove that the points (2, −1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.
If θ is the angle which the straight line joining the points (x1, y1) and (x2, y2) subtends at the origin, prove that \[\tan \theta = \frac{x_2 y_1 - x_1 y_2}{x_1 x_2 + y_1 y_2}\text { and } \cos \theta = \frac{x_1 x_2 + y_1 y_2}{\sqrt{{x_1}^2 + {y_1}^2}\sqrt{{x_2}^2 + {y_2}^2}}\].
Show that the tangent of an angle between the lines \[\frac{x}{a} + \frac{y}{b} = 1 \text { and } \frac{x}{a} - \frac{y}{b} = 1\text { is } \frac{2ab}{a^2 - b^2}\].
The equation of the line with slope −3/2 and which is concurrent with the lines 4x + 3y − 7 = 0 and 8x + 5y − 1 = 0 is
Find k, if the slope of one of the lines given by kx2 + 8xy + y2 = 0 exceeds the slope of the other by 6.
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 ______.
Find the equation of the straight line passing through (1, 2) and perpendicular to the line x + y + 7 = 0.
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.
If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.
The points A(– 2, 1), B(0, 5), C(– 1, 2) are collinear.
The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Then the other two sides are y – 3 = `(2 +- sqrt(3)) (x - 2)`.
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.
Line joining the points (3, – 4) and (– 2, 6) is perpendicular to the line joining the points (–3, 6) and (9, –18).
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 |
