Question

Observe the graph and answer the question that follows:- ​

Answer 1

$\bf \:\large \red{AηsωeR : 1}$

$\tt\implies \:equation \: of \: line \: b \: : \: y \: = - 4$

$\bf \:\large \red{AηsωeR : 2}$

$\tt\implies \:coordinates \: are \: (3, -4)$

$\bf \:\large \red{AηsωeR : 3}$

$\tt\implies \:coordinates \: of \: Q \: = ( - 1, 0)$

$\bf \:\large \red{AηsωeR : 4}$

$\tt\implies \:equation \: of \: line \: a \: : \: x \: = \: 3$

$\bf \:\large \red{AηsωeR : 5}$

$\tt\implies \:line \: n \: has \: equation \: x \: = - 2$

$\bf \:\large \red{AηsωeR : 6}$

$\tt\implies \: line \: n \: and \: line \: a \: \: are \: parallel \: to \: y \: axis$

$\bf \:\large \red{AηsωeR : 7}$

☆ Line 'l' passes through P(3, 2) and Q (-1, 0).

☆ So using two point form of a line, we know that

$\tt\implies \:y-y_{1}=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$

$\tt\implies \:y-0=\dfrac{3 - 0}{2 - ( - 1)}(x + 1)$

$\tt\implies \:y \: = \: x \: + \: 1$

$\bf \:\large \red{AηsωeR : 8}$

$\tt\implies \:coordinates \: of \: P \: is \: (3, 2)$

$\bf \:\large \red{AηsωeR : 9}$

$\tt\implies \:coordinates \: of \: R \: is \: (-2, -4)$

$\bf \:\large \red{AηsωeR : 10}$

$\tt\implies \:equation \: of \: line \: c \: : \: y \: = \: 2$