Find the equation of the straight line given in gradient - 3 and the coordinates of a point on a straight line (2, -5)
Answers
Answer:
If you know the slope
m
(also known as gradient) of a line, and one its points
(
x
0
,
y
0
)
, the equation of the line is given by
y
−
y
0
=
m
(
x
−
x
0
)
In your case,
m
=
−
1
t
, and
(
x
0
,
y
0
)
=
(
t
2
,
2
t
)
. So, the equation is
y
−
2
t
=
−
1
t
(
x
−
t
2
)
we may rewrite this equation as
y
=
−
1
t
x
+
3
t
The other line is
2
y
+
x
=
4
+
2
t
2
, which we can write as
y
=
2
+
t
2
−
x
2
(I just solved for
y
bringing everything else to the right and dividing by
2
).
To find the point where the lines meet, let's ask for both equation to be satisfied. Since we know the expression for
y
for both lines, let's set them to be equal:
−
1
t
x
+
3
t
=
2
+
t
2
−
x
2
Solving for
x
, we get:
−
1
t
x
+
x
2
=
t
2
−
3
t
+
2
x
(
1
2
−
1
t
)
=
t
2
−
3
t
+
2
x
(
t
−
2
2
t
)
=
t
2
−
3
t
+
2
x
=
2
t
t
−
2
(
t
2
−
3
t
+
2
)
The last thing we need to do is to observe that
t
2
−
3
t
+
2
can be factored as
t
2
−
3
t
+
2
=
(
t
−
1
)
(
t
−
2
)
and thus the expression becomes
x
=
2
t
t
−
2
(
t
−
1
)
(
t
−
2
)
=
2
t
(
t
−
1
)
as required.