Question

The line L1 passes through the points A (2, 5) and B (10, 9). The line L2 is parallel to L1 and passes through the origin. The point C lies on L2 such that AC is perpendicular to L2. Find (i) the coordinates of C, (ii) the distance AC.

Answer 1

SOLUTION

GIVEN

The line L1 goes through the points A (2, 5) and B (10, 9). The line L2 is parallel to L1 and goes through the origin. The point C lies on L2 such that AC is perpendicular to L2.

TO DETERMINE

(i) the coordinates of C

(ii) the distance AC.

EVALUATION

Here it is given that the line L1 passes through the points A (2, 5) and B (10, 9).

So the equation of the line L1 is given by

$\displaystyle \sf{ \frac{y - 5}{x - 2} = \frac{9 - 5}{10 - 2} }$

$\displaystyle \sf{ \implies \: \frac{y - 5}{x - 2} = \frac{4}{8} }$

$\displaystyle \sf{ \implies \: \frac{y - 5}{x - 2} = \frac{1}{2} }$

$\displaystyle \sf{ \implies \: x - 2 = 2y - 10 }$

$\displaystyle \sf{ \implies \: x - 2y = - 8 \: \: \: \: - - - (1) }$

Now the line L2 is parallel to L1

So the equation of the line L2 is

$\displaystyle \sf{ \: x - 2y = k \: \: \: \: - - - (2) }$

Now L2 goes through the origin (0,0)

Thus we get k = 0

So Equation of the line L2 is

$\displaystyle \sf{ \: x - 2y = 0 \: \: \: \: - - - (3) }$

Now AC is perpendicular to L2

Thus Equation of AC is

$\displaystyle \sf{ \: 2x + y = c \: \: \: \: - - - (4) }$

Equation 4 goes through the point A(2,5)

Thus we get

$\displaystyle \sf{ \: (2 \times 2) + 5 = c \: \: \: \: }$

$\sf{ \implies \: c = 9}$

So the equation of the line AC is

$\displaystyle \sf{ \: 2x + y = 9 \: \: \: \: - - - (5) }$

Point C is obtained by solving Equation 3 and Equation 5

Solving Equation 3 and Equation 5 we get

$\displaystyle \sf{x = \frac{18}{5} \: \: and \: \: y = \frac{9}{5} }$

Thus the coordinates of C is

$\displaystyle \sf{ \bigg( \frac{18}{5} \:, \: \frac{9}{5} \bigg) }$

The distance AC

$\displaystyle \sf{ = \sqrt{ {\bigg( \frac{18}{5} - 2 \bigg)}^{2} + {\bigg( \frac{9}{5} - 5\bigg)}^{2} }}$

$\displaystyle \sf{ = \sqrt{ {\bigg( \frac{8}{5} \bigg)}^{2} + {\bigg( \frac{16}{5} \bigg)}^{2} }}$

$\displaystyle \sf{ = \sqrt{ \frac{320}{25} }}$

$\displaystyle \sf{ = \frac{8 \sqrt{5} }{5} }$

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