Find the equation of the lines which pass through the point (-2,5) and are equally inclined to the coordinate axes.
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There are two lines possible passing through any point on coordinate plane which are equally inclined to the coordinate axes.
The slopes of the lines are 1 and -1.
let the equation of the required line be y=mx+c
for slope(m) = 1,
so the equation becomes y=x+c
as it passes through (-2,5)
⇒ 5 = -2 + c
⇒ c = 5+2 = 7
So one of the equation is
y = x+7
for slope(m) = -1,
so the equation becomes y=-x+c
as it passes through (-2,5)
⇒ 5 = -(-2) + c
⇒ 5 = 2 + c
⇒ c = 5-2 = 3
So other equation is
y = -x+3
The slopes of the lines are 1 and -1.
let the equation of the required line be y=mx+c
for slope(m) = 1,
so the equation becomes y=x+c
as it passes through (-2,5)
⇒ 5 = -2 + c
⇒ c = 5+2 = 7
So one of the equation is
y = x+7
for slope(m) = -1,
so the equation becomes y=-x+c
as it passes through (-2,5)
⇒ 5 = -(-2) + c
⇒ 5 = 2 + c
⇒ c = 5-2 = 3
So other equation is
y = -x+3
TPS:
if you have any doubt, let me know
yes why m is -1 and 1?
when a line is equally inclined to both axes, the angle it will make with x-axis is 45 degree and -45 degrees. so the slopes(tan 45 and tan(-45)) are 1 and -1
you can understand better if you try to visualize the lines on coordinate plane.....or draw them...
k.i will
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