Find the equation of the line joining A(1, 3) | vedclass

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

Question

(B) Let $P(x, y)$ be any point on the line $AB$. Since the points $A, B,$ and $P$ are collinear,the area of triangle $ABP$ is $0$.Using the determinan

Vedclass · Accepted Answer

$\mp 2$

Vedclass · Answer

(B) Let $P(x, y)$ be any point on the line $AB$. Since the points $A, B,$ and $P$ are collinear,the area of triangle $ABP$ is $0$.Using the determinant formula for the area of a triangle:$\frac{1}{2} \left| \begin{array}{lll} 0 & 0 & 1 \ 1 & 3 & 1 \ x & y & 1 \end{array} ight| = 0$Expanding along the first row:$\frac{1}{2} [1(y - 3x)] = 0 \implies y - 3x = 0 \implies y = 3x$.This is the equation of the line $AB$.Now,given that the area of triangle $ABD$ is $3 \, 	ext{sq units}$,where $A(1, 3), B(0, 0),$ and $D(k, 0)$:$\frac{1}{2} \left| \begin{array}{lll} 1 & 3 & 1 \ 0 & 0 & 1 \ k & 0 & 1 \end{array} ight| = \pm 3$Expanding along the second row:$\frac{1}{2} [(-1) \cdot (0 - 3k)] = \pm 3$$\frac{1}{2} [3k] = \pm 3$$\frac{3k}{2} = \pm 3 \implies 3k = \pm 6 \implies k = \pm 2$.

Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants and find $k$ if $D(k, 0)$ is a point such that the area of triangle $ABD$ is $3 \, \text{sq units}$.

Similar Questions

If the system of equations $ (k+1)^3 x + (k+2)^3 y = (k+3)^3 $, $ (k+1) x + (k+2) y = k+3 $, and $ x + y = 1 $ is consistent, then the value of $ k $ is:

If $\omega$ is an imaginary cube root of unity and $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2\end{array}\right|=0$, then one of the values of $x$ is

If $a_i^2 + b_i^2 + c_i^2 = 1$ for $(i = 1, 2, 3)$ and $a_i a_j + b_i b_j + c_i c_j = 0$ for $(i \ne j, i, j = 1, 2, 3)$,then the value of $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|^2$ is

The value of $x,$ if $\left| \begin{matrix} -x & 1 & 0 \\ 1 & -x & 1 \\ 0 & 1 & -x \end{matrix} \right| = 0$ is equal to

If $A = \begin{bmatrix} 5 & 5x & x \\ 0 & x & 5x \\ 0 & 0 & 5 \end{bmatrix}$ and $|A^2| = 25$, then $|x|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module