Area of Triangle Using Coordinates
Trending Questions
Q. Let P=(1xp, p), Q=(1xq, q) and R=(1xr, r), where xk≠0 denotes the kthterm of an H.P. for k∈N. If the area formed by the points P, Q and R is λpqr sq. units, then the value of λ is
Q. If the mid-points of the sides of a triangle are (1, 1), (2, 4) and (3, 5), then the area of triangle is
(in sq. units)
(in sq. units)
Q. A line L passes through the points (1, 1) and (2, 0) and another line L’ passes through [12, 0] and perpendicular to L. Then the area of the triangle formed by the lines L, L’ and y –axis, is
- 158
- 254
- 258
- 2516
Q. If the co-ordinates of two points A and B are (3, 4) and (5, -2) respectively, find the co-ordinates of any point P if PA=PB and Area of triangle PAB =10
- (2, 7)
- (0, 1)
- (7, 2)
- (1, 0)
Q. If the mid-points of the sides of a triangle are (1, 1), (2, 4) and (3, 5), then the area (in sq. units) of the triangle is
- 5
- 4
- 9
- 10
Q. A line L passes through the points (1, 1) and (2, 0) and another line L’ passes through [12, 0] and perpendicular to L. Then the area of the triangle formed by the lines L, L’ and y –axis, is
- 158
- 254
- 258
- 2516
Q. If the vertices of a triangle are (t, t−2), (t+2, t+2) and (t+3, t), then the area (in sq. units) of the triangle is
- 4
- 8
- 16
- Dependent on the value of t.
Q. A line L passes through the points (1, 1) and (2, 0) and another line L’ passes through [12, 0] and perpendicular to L. Then the area of the triangle formed by the lines L, L’ and y –axis, is
- 158
- 254
- 258
- 2516
Q.
If the area of the triangle formed by the lines y=x, x+y=2 and the line through P(h, k) and parallel to x-axis is 4h2, the locus of P can be
2x−y+1=0
2x+y−1=0
x−2y+1=0
x+2y−1=0
Q.
If the vertices of a triangle be (am21, 2am1), (am22, 2am2) and (am23, 2am3), then the area of the triangle is
a(m2−m3)(m3−m1)(m1−m2)
(m2−m3)(m3−m1)(m1−m2)
a2(m2−m3)(m3−m1)(m1−m2)
4
Q. Let OPQR be a square and M and N be the midpoints of the sides PQ and QR respectively. The ratio of the area of square to the triangle OMN is
- 4:1
- 2:1
- 8:3
- 7:3
Q.
The area of the triangle formed by the intersection of a line parallel to x-axis and passing through P(h, k) with the lines y=x and x+y=2 is 4h2. Find the locus of the point P.
y=2x+1
y=-2x+1
y-2x+1 = 0
2y-x+1 = 0
Q. The points (k, –2k), (–k + 1, 2k) and
(–4 –k, 6 –2k) can't be collinear for any value of k.
(–4 –k, 6 –2k) can't be collinear for any value of k.
- True
- False
Q. The coordinates of points P, Q, R and S are (−3, 5), (4, −2), (p, 3p) and (6, 3) respectively, and the area of △PQR and △QRS are in ratio 2:3, then possible value of p can be
- −543
- 343
- 4641
- 4541
Q. The line 3x + 2y =24 meets y – axis at A and x – axis at B. The perpendicular bisector of AB meets the line through (0, –1) parallel to x – axis at C. The area of the triangle ABC is
- 182sq. Units
- 91 sq. Units
- 48 sq. Units
- 501 sq units
Q. Let A (h, k), B (1, 1) and C (2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which `k' can take is given by
- {1, 3}
- {0, 2}
- {-1, 3}
- {-3, -2}
Q. The points (k, –2k), (–k + 1, 2k) and
(–4 –k, 6 –2k) can't be collinear for any value of k.
(–4 –k, 6 –2k) can't be collinear for any value of k.
- True
- False
Q. Let A (h, k), B (1, 1) and C (2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which `k' can take is given by
- {1, 3}
- {0, 2}
- {-1, 3}
- {-3, -2}
Q. If the vertices of a triangle are (t, t−2), (t+2, t+2) and (t+3, t), then the area (in sq. units) of the triangle is
- 4
- 8
- 16
- Dependent on the value of t.
Q.
Find the ratio in which the line segment joining the points and is divided by . Also find the coordinates of the point of division.
Q. If the vertices of a triangle are (1, 2), (4, −6) and (3, 5), then the area (in sq. units) of the triangle is
- 252
- 12
- 5
- 25
Q. Let PAB be a triangle in which PA=PB and the coordinates of A and B are (−1, −6) and (−5, −2) respectively. If the area of △PAB is 8 square units, then the coordinates of P are
- (−1, −2)
- (−2, −3)
- (−4, −5)
- (−5, −6)
Q. The point A divides the line segment joining the points (−5, 1) and (3, 5) in the ratio k:1. The coordinates of points B and C are (1, 5) and (7, −2) respectively. If the area of △ABC is 2 square units, then the number of values of k is
- 1
- 2
- 3
- 4