Advertisements
Advertisements
प्रश्न
O(0, 0), A(3, 5) and B(−5, −3) are the vertices of triangle OAB. Find the equation of altitude of triangle OAB through vertex B.
उत्तर
The altitude through vertex B is perpendicular to OA.
Slope of OA = `(5 - 0)/( 3 - 0) = 5/3`
Slope of the required altitude = `(-1)/(5/3)= (-3)/5`
The equation of the required altitude through B is
y – y1 = m(x – x1)
`y + 3 = (-3)/5 (x + 5)`
5y + 15 = −3x − 15
3x + 5y + 30 = 0
APPEARS IN
संबंधित प्रश्न
Use graph paper for this question (Take 2 cm = 1 unit along both x and y-axis). ABCD is a quadrilateral whose vertices are A(2, 2), B(2, –2), C(0, –1) and D(0, 1).
1) Reflect quadrilateral ABCD on the y-axis and name it as A'B'CD
2) Write down the coordinates of A' and B'.
3) Name two points which are invariant under the above reflection
4) Name the polygon A'B'CD
Using a graph paper, plot the points A(6, 4) and B(0, 4).
- Reflect A and B in the origin to get the images A' and B'.
- Write the co-ordinates of A' and B'.
- State the geometrical name for the figure ABA'B'.
- Find its perimeter.
A straight line passes through the points P(–1, 4) and Q(5, –2). It intersects the co-ordinate axes at points A and B. M is the mid-point of the segment AB. Find:
- The equation of the line.
- The co-ordinates of A and B.
- The co-ordinates of M.
Show that A(3, 2), B(6, −2) and C(2, −5) can be the vertices of a square.
- Find the co-ordinates of its fourth vertex D, if ABCD is a square.
- Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
O(0, 0), A(3, 5) and B(−5, −3) are the vertices of triangle OAB. Find the equation of median of triangle OAB through vertex O.
A line AB meets the x-axis at point A and y-axis at point B. The point P(−4, −2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:
- the co-ordinates of A and B.
- equation of line through P and perpendicular to AB.
Point A and B have co-ordinates (7, −3) and (1, 9) respectively. Find:
- the slope of AB.
- the equation of perpendicular bisector of the line segment AB.
- the value of ‘p’ of (−2, p) lies on it.
A line is of length 10 units and one end is at the point (2, – 3). If the abscissa of the other end be 10, prove that its ordinate must be 3 or – 9.
Use a graph sheet for this question, take 2 cm = 1 unit along both x and y-axis:
- Plot the points A (3, 2) and B (5, 0). Reflect point A on the y-axis to A΄. Write co-ordinates of A΄.
- Reflect point B on the line AA΄ to B΄. Write the co-ordinates of B΄.
- Name the closed figure A’B’AB.
