The points (-a,-b), (a, b), (0,0) and (a^{2}, | vedclass

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

Question

(A) Let the four points be $A(-a,-b)$, $B(a, b)$, $C(0,0)$, and $D(a^{2}, ab)$.To check if $A, B$, and $C$ are collinear, we calculate the determinant

Vedclass · Accepted Answer

collinear

Vedclass · Answer

(A) Let the four points be $A(-a,-b)$, $B(a, b)$, $C(0,0)$, and $D(a^{2}, ab)$.To check if $A, B$, and $C$ are collinear, we calculate the determinant:$\left|\begin{array}{ccc}-a & -b & 1 \ a & b & 1 \ 0 & 0 & 1\end{array}ight| = -a(b-0) + b(a-0) + 1(0) = -ab + ab = 0$.Since the determinant is $0$, points $A, B$, and $C$ are collinear.Now, check if $B, C$, and $D$ are collinear:$\left|\begin{array}{ccc}a & b & 1 \ 0 & 0 & 1 \ a^{2} & ab & 1\end{array}ight| = a(0-ab) - b(0-a^{2}) + 1(0) = -a^{2}b + a^{2}b = 0$.Since the determinant is $0$, points $B, C$, and $D$ are collinear.Since $A, B, C$ are collinear and $B, C, D$ are collinear, all four points $A, B, C$, and $D$ must lie on the same line.

The points $(-a,-b), (a, b), (0,0)$ and $(a^{2}, ab)$ where $a \neq 0, b \neq 0$ are always

Similar Questions

$A$ vertex of a square is $(3, 4)$ and one diagonal lies on the line $x + 2y = 1$. Find the equation of the second diagonal which passes through the given vertex.

The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equations of the sides $AB$ and $AC$ are $px + qy = 1$ and $qx + py = 1$. The equation of the median through $A$ is:

If the extremities of the base of an isosceles triangle are the points $(2a, 0)$ and $(0, a)$ and the equation of one of the sides is $x = 2a$,then the area of the triangle is

The triangle with vertices at $(-2, 2)$,$(2, -2)$,and $(1, 1)$ is:

The area of the parallelogram formed by the lines $a_1x + b_1y + c_1 = 0$,$a_1x + b_1y + d_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_2x + b_2y + d_2 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module