Advertisements
Advertisements
प्रश्न
Find a relation between x and y, if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear.
उत्तर
Let `A(x_1=x,y_1=y),B(x_2=-5,y_2=7) and C (x_3 =-4, y_3=5)` be the given points
The given points are collinear if
`x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0`
`⇒ x(7-5)+(-5)(5-y)+(-4)(y-7)=0`
`⇒7x-5x-25+5y-4y+28=0`
`⇒ 2x +y+3=0`
Hence, the required relation is 2x +y+3 =0
APPEARS IN
संबंधित प्रश्न
In each of the following find the value of 'k', for which the points are collinear.
(8, 1), (k, -4), (2, -5)
If the coordinates of the mid-points of the sides of a triangle are (3, 4) (4, 6) and (5, 7), find its vertices.
Show that the following points are collinear:
A(5,1), B(1, -1) and C(11, 4)
Show that ∆ ABC with vertices A (–2, 0), B (0, 2) and C (2, 0) is similar to ∆ DEF with vertices D (–4, 0), F (4, 0) and E (0, 4) ?
Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear ?
If the points A (x, y), B (3, 6) and C (−3, 4) are collinear, show that x − 3y + 15 = 0.
In Figure 1, PS = 3 cm, QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, PQ ⊥ RQ and RQ = 9 cm. Evaluate tan θ.
Show that the ∆ABC is an isosceles triangle if the determinant
Δ = `[(1, 1, 1),(1 + cos"A", 1 + cos"B", 1 + cos"C"),(cos^2"A" + cos"A", cos^2"B" + cos"B", cos^2"C" + cos"C")]` = 0
The area of the triangle whose vertices are A(1, 2), B(-2, 3) and C(-3, -4) is ______.
The area of ∆ABC is 8 cm2 in which AB = AC = 4 cm and ∠A = 90º.