Advertisements
Advertisements
प्रश्न
If two opposite vertices of a square are (1, 2) and (5, 8), find the coordinates of its other two vertices and the equations of its sides.
Advertisements
उत्तर
Slope of AC = \[\frac{8 - 2}{5 - 1} = \frac{3}{2}\]
The sides AB and AD pass through the point A(1,2) and make an angle of \[{45}^\circ\] with AC whose slope is \[\frac{3}{2}\].
Equations of AB and AD are given by \[y - 2 = \frac{\frac{3}{2} \pm \tan {45}^\circ}{1 \mp \frac{3}{2}\tan {45}^\circ}\left( x - 1 \right)\]
\[\Rightarrow y - 2 = \frac{3 \pm 2}{2 \mp 3}\left( x - 1 \right)\]
\[\Rightarrow y - 2 = - 5\left( x - 1 \right) \text { and } y - 2 = \frac{1}{5}\left( x - 1 \right)\]
\[ \Rightarrow 5x + y - 7 = 0 \text { and } x - 5y + 9 = 0\]
Thus, the equations of AB and AD are \[5x + y - 7 = 0 \text { and } x - 5y + 9 = 0\] respectively.
Since BC is parallel to AD, the equation of BC is \[x - 5y + \lambda = 0\].
This line passes through C (5,8).
\[5 - 40 + \lambda = 0 \Rightarrow \lambda = 35\]
So, the equation of BC is \[x - 5y + 35 = 0\].
Since CD is parallel to AB, the equation of CD is \[5x + y + \lambda = 0\].
This line passes through C (5, 8).
\[25 + 8 + \lambda = 0 \Rightarrow \lambda = - 33\]
So, the equation of CD is \[5x + y - 33 = 0\].
Solving equation of AB and BC, we get B as (0, 7).
Solving equation of AD and CD, we get D as (6, 3).
Hence, the other two vertices are (0, 7) and (6, 3).
APPEARS IN
संबंधित प्रश्न
The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.
Find the slope of the line, 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 can be said regarding a line if its slope is zero ?
What can be said regarding a line if its slope is negative?
Show that the line joining (2, −5) and (−2, 5) is perpendicular to the line joining (6, 3) and (1, 1).
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.
Consider the following population and year graph:
Find the slope of the line AB and using it, find what will be the population in the year 2010.
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.
A quadrilateral has vertices (4, 1), (1, 7), (−6, 0) and (−1, −9). Show that the mid-points of the sides of this quadrilateral form a parallelogram.
Find the equation of a straight line with slope −2 and intersecting the x-axis at a distance of 3 units to the left of origin.
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.
Find the equation of the strainght line intersecting y-axis at a distance of 2 units above the origin and making an angle of 30° with the positive direction of the x-axis.
Find the equations of the altitudes of a ∆ ABC whose vertices are A (1, 4), B (−3, 2) and C (−5, −3).
Find the angles between the following pair of straight lines:
(m2 − mn) y = (mn + n2) x + n3 and (mn + m2) y = (mn − n2) x + m3.
Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.
Show that the tangent of an angle between the lines \[\frac{x}{a} + \frac{y}{b} = 1 \text { and } \frac{x}{a} - \frac{y}{b} = 1\text { is } \frac{2ab}{a^2 - b^2}\].
Write the coordinates of the image of the point (3, 8) in the line x + 3y − 7 = 0.
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
The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y − 11 = 0 are
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 ______.
A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5, 3). Find the coordinates of 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 a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of x-axis.
A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P.
The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is ______.
Equation of the line passing through (1, 2) and parallel to the line y = 3x – 1 is ______.
The point (4, 1) undergoes the following two successive transformations:
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the positive x-axis Then the final coordinates of the point are ______.
Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are ______.
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.
The lines whose vector equations are `r = 2hati - 3hatj + 7hatk + lambda (2hati + phatj + 5hatk) and r = hati - 2hatj + 3hatk + µ(3hati + phatj + phatk)` are perpendicular for all values of λ and µ if p =
