Advertisements
Advertisements
Question
If the points A(−1, −4), B(b, c) and C(5, −1) are collinear and 2b + c = 4, find the values of b and c.
Solution
As the given points are collinear, so the area of the triangle formed by them must be 0.
∴ 1/2[x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)] = 0
Here, x1= −1 , y1= −4, x2= b, y2= c, x3= 5, y3= −1
∴12[−1(c+1)+b(−1+4)+5(−4−c)]=0
⇒−c−1−b+4b−20−5c=0
⇒−6c+3b−21=0
⇒b−2c=7 ...(1)
Given:
2b+c=4 ...(2)
On solving equation (1) and (2), we get:
b=3 and c=−2
APPEARS IN
RELATED QUESTIONS
If the points A(x, 2), B(−3, −4) and C(7, − 5) are collinear, then the value of x is:
(A) −63
(B) 63
(C) 60
(D) −60
The two opposite vertices of a square are (− 1, 2) and (3, 2). Find the coordinates of the other two vertices.
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 points (-3, -3),(3,3) and C (-3 `sqrt(3) , 3 sqrt(3))` are the vertices of an equilateral triangle.
Show that the points A(-5,6), B(3,0) and C(9,8) are the vertices of an isosceles right-angled triangle. Calculate its area.
If the centroid of ΔABC having vertices A (a,b) , B (b,c) and C (c,a) is the origin, then find the value of (a+b+c).
If the area of triangle ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove that x + y = 15 ?
Show that the points (a + 5, a – 4), (a – 2, a + 3) and (a, a) do not lie on a straight line for any value of a.
Let `Delta = abs (("x", "y", "z"),("x"^2, "y"^2, "z"^2),("x"^3, "y"^3, "z"^3)),` then the value of `Delta` is ____________.
Area of a triangle = `1/2` base × ______.