Question

find the points on the line x+y=5 that lie at the distance 2unit from the line 4x+3y -12=0 ​​

Answer 1

Answer:

ANSWER

Note that the co-ordinates of an arbitrary point on x + y = 4 can be obtained by putting x = t (or y = t) and then obtaining y (or x) from the equation of the line where t is a parameter

Putting x = t in the equation x + y = 4 of the given line we obtain y = 4 - t So co-ordinates of an arbitrary point on the given line are P(t, 4 - t) Let P(t, 4 - t) be the required point Then distance of P from the line 4x + 3y - 10 = 0 is unity i.e.

⇒

∣

∣

∣

∣

∣

4

2

+3

2

4t+3(4−t)−10

∣

∣

∣

∣

∣

=1⇒∣t+2∣=5

⇒t+2=±5 ⇒ t=−7 or t=3.

Hence the required points are (−7,11) or (3,1)

Answer 2

Step-by-step explanation:

Note that the co-ordinates of an arbitrary point on x + y = 4 can be obtained by putting x = t (or y = t) and then obtaining y (or x) from the equation of the line where t is a parameter

Putting x = t in the equation x + y = 4 of the given line we obtain y = 4 - t So co-ordinates of an arbitrary point on the given line are P(t, 4 - t) Let P(t, 4 - t) be the required point Then distance of P from the line 4x + 3y - 10 = 0 is unity i.e.

⇒

∣

∣

∣

∣

∣

4

2

+3

2

4t+3(4−t)−10

∣

∣

∣

∣

∣

=1⇒∣t+2∣=5

⇒t+2=±5 ⇒ t=−7 or t=3.

Hence the required points are (−7,11) or (3,1)