JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 2)
Let L denote the line in the xy-plane with x and
y intercepts as 3 and 1 respectively. Then the
image of the point (–1, –4) in this line is :
$$\left( {{{11} \over 5},{{28} \over 5}} \right)$$
$$\left( {{{29} \over 5},{{11} \over 5}} \right)$$
$$\left( {{{29} \over 5},{8 \over 5}} \right)$$
$$\left( {{8 \over 5},{{29} \over 5}} \right)$$
Explanation
Line is $${x \over 3} + {y \over 1} = 1$$
$$ \Rightarrow $$ x + 3y – 3 = 0
Let Image of point (–1, –4) is ($$\alpha $$, $$\beta $$)
Hence, $${{\alpha + 1} \over 1} = {{\beta + 4} \over 3} = - 2\left( {{{ - 1 - 12 - 3} \over {10}}} \right)$$
$$ \Rightarrow $$ $${{\alpha + 1} \over 1} = {{\beta + 4} \over 3} = {{16} \over 5}$$
$$ \Rightarrow $$ $$\alpha $$ = $${{11} \over 5}$$, $$\beta $$ = $${{28} \over 5}$$
$$ \Rightarrow $$ x + 3y – 3 = 0
Let Image of point (–1, –4) is ($$\alpha $$, $$\beta $$)
Hence, $${{\alpha + 1} \over 1} = {{\beta + 4} \over 3} = - 2\left( {{{ - 1 - 12 - 3} \over {10}}} \right)$$
$$ \Rightarrow $$ $${{\alpha + 1} \over 1} = {{\beta + 4} \over 3} = {{16} \over 5}$$
$$ \Rightarrow $$ $$\alpha $$ = $${{11} \over 5}$$, $$\beta $$ = $${{28} \over 5}$$
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