Question

The distance of the line 2x - 3y = 4 from the point (1, 1) measured parallel to the line x + y = 1 is​

Answer 1

Answer:

The distance of the lines 2x - 3y = 4 from the point (1, 1) measured parallel to the line x + y = 1 is. The point of intersection is (2, 0). Hence option (1) is the answer

Answer 2

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \bf \to \: given \: : \\ \\ \rm \: 2x - 3y = 4 \: \: \: - - - (1) \\ \\ \rm \: x + y = 1 \: \: \: \: - - - - (2) \\ \\ \sf \: let \: assume \: that \\ \\ \rm \: the \: equation \: of \: line \: parallel \: to \: (2) \\ \rm \: will \: be \: \: \\ \\ \rm \: x + y + k = 0 \: \: \: - - - - (3) \\ \\ \sf \: again \: given \: that \\ \\ \rm \: eqn(3) \: passes \: through \: the \: point \: (1,1) \\ \\ \sf \: put \: this \: in \: eqn.(3) = > \\ \\ \rm \: 1 + 1 + k = 0 \\ \\ = > \rm \: k = - 2 \\ \\ \sf \: now \: eqn.(3) \: can \: be \: written \: as \\ \\ \rm \: x + y - 2 = 0 \: \: \: \: \: - - - (4) \\ \\ \sf \: now \: we \: have \: to \: solve \: eqn.(1) \: and \: eqn.(4) \\ \\ \sf \: from \: eqn.(4) = > \\ \rm \: y = 2 - x \: \: \: - - - (5) \\ \sf \: substitute \: this \: in \: eqn.(1) = > \\ \\ \rm \: 2x - 3(2 - x) = 4 \\ \\ = > 2x - 6 + 3x = 4 \\ \\ = > 5x = 10 \\ \\ = > x = 2 \\ \\ \sf \: put \: this \: in \: eqn.(5) = > \\ \\ y = 2 - 2 \\ \\ = > y = 0 \\ \\ \therefore \sf \: intersection \: point \: of \: eqn.(1) \: and \: eqn.(4) \\ \sf \: is \: (2,0) \\ \\ \\ \sf \: now \: distance \: between \: (1,1) \: and \: (2,0) \: is \: \\ \\ \rm \: d = \sqrt{ {(1 - 2)}^{2} + {(1 - 0)}^{2} } \\ \\ = \sqrt{ {( - 1)}^{2} + {1}^{2} } \\ \\ = \sqrt{1 + 1} \\ \\ = \sqrt{2 } \: unit \\ \\ \rm = \: answer \end{array}}}}}$