- Find a point on the y – axis which is equidistant from the point A (6, 5) and B (-4, 3).
Solution
Let the point of Y-axis is (0,y)
Then
$\sqrt{(0-6)^{2}+(y-5)^{2}}=\sqrt{(0+4)^{2}+(y-3)^{2}}$
Squaring both the sides
${(0-6)^{2}+(y-5)^{2}}={(0+4)^{2}+(y-3)^{2}}$
36+y^{2}+25-10y=16+y^{2}+9-6y
4y=36
y=9
So point (0,9)
- Show that the points (p, p), (-p, -p) and (-p√3, p√3) are the vertices of an equilateral triangle. Also find its area.
Solution
Let the point be A(p,p) ,B(-p,-p) and C(-p√3, p√3)
Then
AB=$\sqrt{(-p-p)^{2}+(-p-p)^{2}}$=2p√2
BC=$\sqrt{(-p\sqrt{3}+p)^{2}+(p\sqrt{3}+p)^{2}}$=2p√2
AC=$\sqrt{(-p\sqrt{3}-p)^{2}+(p\sqrt{3}-p)^{2}}$=2p√2
All Sides equal,It is an equilateral triangle
Area of triangle
=$\frac{(side)^2\sqrt {3}}{4}$
=2 √3p^{2} sq. units
- Check whether the points (4, 5), (7, 6) and (6, 3) are collinear
Solution
If the point are collinear, then Area will be zero
$A=\frac{1}{2}[4(6-3)+7(3-5)+6(5-6)]=-8$
So they are not collinear
- Find the value of q for which the points (7, -2), (5, 1), and (3, q) are collinear.
Solution
If the point are collinear, then Area must be zero
$0=\frac{1}{2}[7(1-q)+5(q+2)+3(-2-1)]$
q=4
- Show that four points (0, -1), (6, 7), (-2, 3) and (8, 3) are the vertices of a rectangle. Also find its area.
Solution
Let the point be A(0,-1) ,B(6,7) , C(-2, 3) and D(8,3)3)
Then
AB=$\sqrt{(0-6)^{2}+(-1-7)^{2}}$=10
BC=$\sqrt{(6+2)^{2}+(7-3)^{2}}$=2√20
CD=$\sqrt{(-2-8)^{2}+(3-3)^{2}}$=10
AD=$\sqrt{(0-8)^{2}+(3+1)^{2}}$=2√20
AC=$\sqrt{(0+2)^{2}+(-1-3)^{2}}$=√20
BD=$\sqrt{(6-8)^{2}+(7-3)^{2}}$=√20
So Opposite sides are equal
BC=AB
AC=BD
Diagonals are equal
AB= CD
So these are coordinates of the rectanges
Now Area = Length X breath
=2√20 X √20=40
- If two vertices of an equilateral triangle be (0,0), (3, √3), find the third vertex.
Solution
Two vertices of an equilateral triangle are (0, 0) and (3, √3).
Let the third vertex of the equilaterla triangle be (x, y)
Distance between (0, 0) and (x, y) = Distance between (0, 0) and (3, √3) = Distance between (x, y) and (3, √3)
√(x^{2}+ y^{2}) = √(3^{2}); + 3) = √[(x - 3)^{2} + (y - √3)^{2}]
x^{2} + y^{2} = 12 --(A)
√(x^{2} + y^{2})=√[(x - 3)^{2} + (y - √3)^{2}]
x^{2} + 9 - 6x + y^{2} + 3 - 2√3y = x^{2} + y^{2}
12 - 6x - 2√3y = 0
3x + √3y = 6
x = (6 - √3y) / 3 -(B)
Substituting the values of x in (A)
[(6 - √3y)/3]^{2} + y^{2} = 12
(36 + 3y^{2} - 12√3y) / 9 + y^{2} = 12
36 + 3y^{2} - 12√3y + 9y^{2} = 108
- 12√3y + 12y^{2} - 72 = 0
-√3y + y^{2} - 6 = 0
(y - 2√3)(y + √3) = 0
y = 2√3 or - √3
If y = 2√3, x = (6 - 6) / 3 = 0
If y = -√3, x = (6 + 3) / 3 = 3
So, the third vertex of the equilateral triangle = (0, 2√3) or (3, -√3).
- If P (2, -1), Q (3, 4), R (-2, 3) and S (-3, -2) be four points is a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.
- Show that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices points of a rectangle.
- Show that the points A (1, -2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.
- Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square.
- Prove that the points (3, 0), (6, 4) and (-1, 3) are vertices of a right angled isosceles triangle.
- Find the circumcenter of the triangle whose vertices are (-2, -3), (-1, 0), (12, -6)
- If two opposite vertices of a square are (5, 4) and (1, -6), find the co- ordinates of its remaining two vertices.
- Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.
- Find the value of x such that PQ = QR where the co- ordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.