RD Chapter 14 Co ordinate Geometry Ex 14.2 Solutions
Question - 11 : - Prove that the points (2a, 4a), (2a, 6a) and (2a + √3 a , 5a) are the vertices of an equilateral triangle.
Answer - 11 : -
Question - 12 : - Prove that the points (2, 3), (-4, -6) and (1, 3/2 )do not form a triangle.
Answer - 12 : -
Question - 13 : - The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC. [NCERT Exemplar]
Answer - 13 : - Given that, the points A (2, 9), B (a, 5) and C (5, 5) are the verticesof a ∆ABC right angled at B.
By Pythagoras theorem, AC² = AB² + BC² ………(i)
Now, by distance formula,
Question - 14 : - Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2, 3) and (-3, -2) is a rhombus.
Answer - 14 : -
Question - 15 : - Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.
Answer - 15 : - Two vertices of an isosceles ∆ABC are A (2, 0) and B (2, 5). Let co-ordinates of third vertex C be (x, y)
Question - 16 : - Which point on x-axis is equidistant from (5, 9) and (-4, 6) ?
Answer - 16 : -
Let co-ordinates of two points are A (5, 9), B (-4, 6)
The required point is on x-axis
Its ordinates or y-co-ordinates will be 0
Let the co-ordinates of the point C be (x, 0)
AC = CB
Question - 17 : - Prove that the point (-2, 5), (0, 1) and (2, -3) are collinear.
Answer - 17 : -
Now AB + BC = 2√5 +2√5
and CA = 4√5
AB + BC = CA
A, B and C are collinear
Question - 18 : - The co-ordinates of the point P are (-3,2). Find the co-ordinates of the point Q which lies on the line joining P and origin such that OP = OQ.
Answer - 18 : -
Co-ordinates of P are (-3, 2) and origin O are (0, 0)
Let co-ordinates of Q be (x, y)
O is the midpoint of PQ
= 9 + 4 = (±3)² + (±2)²
The point will be in fourth quadrant
Its y-coordinates will be negative
and x-coordinates will be positive
Now comparing the equation
x² = (±3)² => x = ±3
y² = (±2)² => y = ±2
x = 3, y = -2
Co-ordinates of the point Q are (3, -2)
Question - 19 : - Which point on y-axis is equidistant from (2, 3) and (-4, 1) ?
Answer - 19 : -
The required point lies on y-axis
Its abscissa will be zero
Let the point be C (0, y) and A (2, 3), B (-4, 1)
Now,
Question - 20 : - The three vertices of a parallelogram are (3, 4), (3, 8) and (9, 8). Find the fourth vertex.
Answer - 20 : - Let ABCD be a parallelogram and vertices will be A (3, 4), B (3, 8), C (9, 8)