Advertisements
Advertisements
प्रश्न
Find the value of p for which the points (−5, 1), (1, p) and (4, −2) are collinear.
उत्तर
Points are collinear means the area of the triangle formed by the collinear points is 0.
Using the area of a triangle = `[ 1/2 [x_1(y_2 − y_3) + x_2( y_3 − y_1) + x_3(y_1 − y_2)]`
=`1/2[ −5(p − ( −2)) + 1( −2 − 1) + 4( 1− p)]`
= `1/2[ −5( p + 2) + 1( −3 ) + 4(1 − p)]`
= `1/2[ −5p − 10 − 3 + 4 − 4p]`
= `1/2[ − 5p − 9 − 4p ]`
Area of triangle will be zero points being collinear
`1/2[ −5p − 4p − 9 ]`=0
`1/2[ −9p − 9 ] = 0`
9p + 9 = 0
p = − 1
Therefore, the value of p = −1.
संबंधित प्रश्न
If D, E and F are the mid-points of sides BC, CA and AB respectively of a ∆ABC, then using coordinate geometry prove that Area of ∆DEF = `\frac { 1 }{ 4 } "(Area of ∆ABC)"`
Find the area of the quadrilateral ABCD whose vertices are respectively A(1, 1), B(7, –3), C(12, 2) and D(7, 21).
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).
Prove that the points (2a, 4a), (2a, 6a) and `(2a + sqrt3a, 5a)` are the vertices of an equilateral triangle.
If `a≠ b ≠ c`, prove that the points (a, a2), (b, b2), (c, c2) can never be collinear.
Four points A (6, 3), B (−3, 5), C(4, −2) and D (x, 3x) are given in such a way that `(ΔDBG) /(ΔABG)=1/2,` find x
Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the coordinates of the third vertex.
Show that the points O(0,0), A`( 3,sqrt(3)) and B (3,-sqrt(3))` are the vertices of an equilateral triangle. Find the area of this triangle.
If the points A (x, y), B (3, 6) and C (−3, 4) are collinear, show that x − 3y + 15 = 0.
If the points (a1, b1), (a2, b2) and(a1 + a2, b1 + b2) are collinear, then ____________.