Find the value of k for which the points (8,1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For points to be collinear,the area of the triangle formed by them must be $0$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For points to be collinear,the area of the triangle formed by them must be $0$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.Substituting the given points $(8,1), (k,-4),$ and $(2,-5)$ into the formula:$\frac{1}{2} |8(-4 - (-5)) + k(-5 - 1) + 2(1 - (-4))| = 0$$\frac{1}{2} |8(-4 + 5) + k(-6) + 2(1 + 4)| = 0$$\frac{1}{2} |8(1) - 6k + 2(5)| = 0$$|8 - 6k + 10| = 0$$18 - 6k = 0$$6k = 18$$k = 3$

Find the value of $k$ for which the points $(8,1), (k,-4),$ and $(2,-5)$ are collinear.

Similar Questions

Find the distance between the following pairs of points: $(-5, 7)$ and $(-1, 3)$. (in $\sqrt{2}$)

Find the area of a triangle whose vertices are $(1, -1), (-4, 6)$ and $(-3, -5)$ (in square units).

$A$ median of a triangle divides it into two triangles of equal areas. Verify this result for $\Delta ABC$ whose vertices are $A (4, -6), B (3, -2)$ and $C (5, 2)$. Find the area of each triangle (in square units).

Find the area of the triangle whose vertices are $(2, 3), (-1, 0),$ and $(2, -4)$.

Let $A(4,2)$,$B(6,5)$,and $C(1,4)$ be the vertices of $\Delta ABC$. The median from $A$ meets $BC$ at $D$. Find the coordinates of the point $D$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module