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Question

The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC.

If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q.


Solution

For the line AB:

S l o p e space o f space A B equals m equals fraction numerator 2 minus 0 over denominator 2 minus 4 end fraction equals fraction numerator 2 over denominator negative 2 end fraction equals negative 1

(x1, y1) = (4, 0)

Equation of the line AB is

y - y1 = m(x - x1)

y - 0 = -1(x - 4)

y = -x + 4

x + y = 4 ....(1)

For the line BC:

S l o p e space o f space B C equals m equals fraction numerator 6 minus 2 over denominator 0 minus 2 end fraction equals fraction numerator 4 over denominator negative 2 end fraction equals negative 2

(x1, y1) = (2, 2)

Equation of the line BC is

y - y1 = m(x - x1)

y - 2 = -2(x - 2)

y - 2 = -2x + 4

2x + y = 6 ....(2)

Given that AB cuts the y-axis at P. So, the abscissa of point P is 0.

Putting x = 0 in (1), we get,

y = 4

Thus, the coordinates of point P are (0, 4).

Given that BC cuts the x-axis at Q. So, the ordinate of point Q is 0.

Putting y = 0 in (2), we get,

2x = 6  x = 3

Thus, the coordinates of point Q are (3, 0).

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