The point of intersection of the lines joinin | vedclass

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(A) Let the points be $A(1, 2)$,$B(2, -1)$,$C(-1, 0)$,and $D(2, 0)$.The equation of the line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$

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$\frac{5}{3} \hat{i}$

Vedclass · Answer

(A) Let the points be $A(1, 2)$,$B(2, -1)$,$C(-1, 0)$,and $D(2, 0)$.The equation of the line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(y - y_1) = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.For line $AB$ passing through $(1, 2)$ and $(2, -1)$:$(y - 2) = \frac{-1 - 2}{2 - 1}(x - 1) \Rightarrow (y - 2) = -3(x - 1) \Rightarrow y - 2 = -3x + 3 \Rightarrow 3x + y = 5$.For line $CD$ passing through $(-1, 0)$ and $(2, 0)$:Since the $y$-coordinates are both $0$,the line is the $x$-axis,which is $y = 0$.To find the intersection,substitute $y = 0$ into the equation $3x + y = 5$:$3x + 0 = 5 \Rightarrow x = \frac{5}{3}$.Thus,the point of intersection is $(\frac{5}{3}, 0)$,which in vector form is $\frac{5}{3} \hat{i}$.

The point of intersection of the lines joining points $\hat{i}+2 \hat{j}, 2 \hat{i}-\hat{j}$ and $-\hat{i}, 2 \hat{i}$ is

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