Name: Properties of Determinants (Part 3)
Uploaded: 2019-07-19
Duration: 3 min 53 s
Description: Suppose we multiply a row of a determinant with a particular value, then what would be the change in the value of the determinant? Watch this video to learn...

Question

Using determinants, find the equation of the line passing through the points $(1, 2)$ and $(3, 6)$ .

Accepted Answer

Given,

Let $P (x, y)$ be any point on the line $AB$ .

$∴$ Points $A, P$ and $B$ are collinear

$∴$ Coordinates of vertices of triangle are

$A (1, 2)$ , $P (x, y)$ , $B (3, 6)$

Area of triangle $APB$ is the absolute value of determinant, given by

Area of triangle $= \frac{1}{2} |\begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix}| = 0 ∴$ Points are collinear

$= \frac{1}{2} |\begin{matrix} 1 & 2 & 1 \\ x & y & 1 \\ 3 & 6 & 1 \end{matrix}| = 0$

Using definition of expansion of determinants,we get

$= \frac{1}{2} [1 |\begin{matrix} y & 1 \\ 6 & 1 \end{matrix}| - 2 |\begin{matrix} x & 1 \\ 3 & 1 \end{matrix}| + 1 |\begin{matrix} x & y \\ 3 & 6 \end{matrix}|] = 0$

Cross multiply by $\frac{1}{2}$ , we get

$= [1 (y \times 1 - 6 \times 1) - 2 (x \times 1 - 3 \times 1) + 1 (x \times 6 - 3 \times y)] = 0$

$= [1 (y - 6) - 2 (x - 3) + (6 x - 3 y)] = 0$

$= [y - 6 - 2 x - 3 + 6 x - 3 y] = 0$

$= [4 x - 2 y] = 0$

$= 2 x - y = 0$

Hence, the equation of line joining two point

$2 x - y = 0$ .

Using determinants, find the equation of the line passing through the points $(1, 2)$ and $(3, 6)$ .

Important Questions on Determinants

Learn and Prepare for any exam you want !

Using determinants, find the equation of the line passing through the points 1, 2 and 3, 6.

Important Questions on Determinants

Learn and Prepare for any exam you want !

Using determinants, find the equation of the line passing through the points $(1, 2)$ and $(3, 6)$ .